The Pokemon Classroom: The Sciences
by ajnrules
Summary: I wrote this series to help study for the MCAT. I originally wrote it for fun, but now I'm going to take the MCAT again, so this is now a series. Basically, as the title implies, the Pokemon characters learn about the physical and biological sciences.
1. Chapter 1: Mathematic Concepts

DISCLAIMER: As I said in the description, this series was not written for aesthetic purposes, but to help me study for the MCAT. So don't expect epic character development, a stirring plot, or an interesting read. I also made up most of the last names of the characters, which were all taken from the first season of Pokémon, but I don't own the characters, Nintendo and Shogakukan does. I'd like to thank those two companies for making Pokémon, and also the Kaplan book for most of the science reviews. Also, a lot of the symbols like the equal sign or any of the superscripts don't appear, so this story is going to be pretty darned confusing.

--

_Brrring! Brrring! Brrring!_

The school bell rang clearly and loudly, and the students began streaming into the classroom. Some of them were bright and chipper, while others seemed less than excited to be there.

"I don't see why we have to go to school," said young Ash Ketchum, flanked by his good friends Misty and Brock. "There's nothing that we can learn here that would help us to become Pokémon Masters."

"Well, Professor Oak did seem emphatic about it when he enrolled us. If he supports education, then it must be very important," Misty replied.

"Yeah? Well I don't like it one bit," said Ash.

They went and got a seat near in the middle of the classroom, where a wise but energetic man with a white lab coat stood near in front of a large, whiteboard.

"You're going to be our teacher, Professor Oak?" asked Brock.

"That's right. In order to aid in my study of Pokémon, I've made it my goal to get a good grasp on the physical and biological sciences. These concepts are very important to understand if you want to be a good Pokémon trainer."

"I don't believe it. Why do you need to know about science to become a Pokémon Master? It's not like you have to use physical equations in the middle of a battle," said Ash.

"It's a good thing, because if that's the case, you wouldn't even last a day as a Pokémon trainer," an obnoxious voice called out from behind.

"What? Gary? You're in this class too?" Ash cried.

"That's right, loser. You don't think that I'd miss a class taught by my own grandfather, do you? This way, I'll be ready when the Pokémon League institutes that rule about doing physics problems in the middle of battle."

"Are they really going to do it?" asked Brock.

"You'd never know."

"All right, Gary, that's enough. And I wish you wouldn't keep your cheerleaders around all the time, but I suppose they can stay as long as they're quiet," Professor Oak said demurely.

"You hear me, ladies?" called Gary. "Now you can watch while I dazzle the crowd in the classroom as well as the Pokémon battlefield!" His cheerleaders cheered.

"Oh, that Gary," Ash murmured bitterly. "I'm not allowed to bring Pikachu, but he can bring those annoying cheerleaders."

"But those cheerleaders are much more pleasant to the eyes than Pikachu is," said Brock.

"All right now, everybody calm down," said Professor Oak. "It's time for us to get started. First, let's have a roll call. Duplica Imite."

"I'm here," said Duplica."

"Ash Ketchum."

"Yeah, yeah. Here," said Ash.

"Lara Laramie."

"Present and accounted for."

"Samurai Miyamoto."

"I am currently in this classroom and ready to learn."

"Gary Oak."

"Yep! I'm present!"

"James Rocket." Silence.

"Okay, Jessie Rocket." More silence.

"Hmm. Meowth Rocket." Even more silence.

"Uh oh," thought Ash. "They can't be in this class too, can they?"

"Okay. I guess we'll give them a few more minutes. Next we have Tracey Sketchit."

"I'm here, and it's certainly great to have you as my professor," said Tracey.

"It's good to have you here too. Brock Slate."

"I'm here."

"Joe Smith."

"Here."

"Todd Snap."

"I'm here, and I can take the class picture once we're done."

"Okay, but no pictures in class. Next is A.J. Stone."

"Yeah, I'm here."

"Richie Taylor."

"I'm ready."

"And finally, Misty Waterflower."

"Present," said Misty.

"All right," said Professor Oak. "Well, nobody came in during that time so it appears as though James, Jessie, and Meowth Rocket decided not to come after all. Well, I'm glad to see all of you that are here. This class–"

A loud crash interrupted Professor Oak, as three bodies came crashing down from the ceiling. "I knew it! It's you!" cried Ash.

The three mysterious people dusted themselves off, and jumped onto the desk.

"Prepare for trouble!"

"And make it double!"

"To protect the world from devastation!"

"To unite all people within our nation!"

"To denounce the evils of truth and love!"

"To extend our reach to the stars above!"

"Jessie!"

"James!"

"Team Rocket blasts off at the speed of light!"

"Surrender now, or prepare to fight!"

"Meowth! That's right!"

The entire class stared at them silently, surprised at their sudden entrance and uncouth interruption of Professor Oak's opening statements. Only the professor seemed glad to see the trio. "Ah, you must be James, Jessie, and Meowth Rocket! I must admit, I didn't think the room would be so hard to find that you'd have to go creeping around in the ceiling, but it's good to have you join us in the endeavor of learning."

"Who's here to learn?" griped Meowth. "We're here to get Pikachu!"

"You're not getting him!" cried Ash. "And besides, I wasn't allowed to bring him!"

"What?" Jessie exclaimed. "This is all your fault, Meowth!"

"Yeah, but at least nobody would be guarding him," said Meowth.

Team Rocket made a run for the door, but to their dismay, the door was locked. "I really do apologize, but the door locks automatically during class for security purposes. Only a pass can unlock the door, but please, your presence in this class would be welcomed."

"But they're Team Rocket! They're here to steal Pokémon!" Ash objected.

"It doesn't matter," Professor Oak replied. "Nobody should be denied the right to an education no matter who they are."

"I don't remember ever signing up for this stupid class," Jessie groused.

"And why did they use a stupid name like James Rocket?" whined James. "I don't like it. It makes it seem like we're a family."

"Well, you kind of are a dysfunctional family," Misty said under her breath.

"Actually, Giovanni, the Viridian City Gym Leader himself signed the three of you up. You should be honored to be associated with such a distinguished individual."

Team Rocket grudgingly sat down as Professor Oak started over on his introduction. "As I was saying, it's good to see all of you here. This class will give you an in-depth overview of the physical and biological sciences. Some of you may be wondering why you're here. Well, I strongly believe that these sciences are very useful in the world of Pokémon training. It's a shame that so many generations of Pokémon trainers have gone without this basic education, but after several years of planning, this Pokémon Classroom is ready to begin. You are all inaugural students in this program."

"So in other words, we're guinea pigs," said A.J.

Some people laughed, but Professor Oak was serious. "You might say so, but I have worked hard with several highly respected scientists and Pokémon researchers to hammer out the curriculum, and I believe that we are ready to teach you what every Pokémon trainer should know about the sciences." He went around and passed out a syllabus. "The four basic physical and biological sciences are Physics, Chemistry, Organic Chemistry, and Biology. All of them are essential to a strong understanding of Pokémon as well as the world around us.

"Even if you don't plan on being a Pokémon trainer, these basic sciences would be very useful. For example, a good understanding of genetics and nutrition would be helpful in being a good Pokémon breeder." Brock nodded. "And an understanding of light and optics would be useful for a cameraman. Due to limitations in time, we would only be able to barely scratch the surface of these sciences, but I hope that we would be able to set a strong foundation for you to build on in your own free time.

"We will begin our study with a look at the physical sciences, physics and chemistry. These two topics are closely tied together so hopefully you can use your understanding of one to help you understand the other. After that, we will transition to the biological sciences, organic chemistry and biology. About grading, we will have regular tests and a final exam, but you shouldn't be so stressed out about these things. If you work hard to build up your knowledge of these underlying concepts, then the tests should be a piece of cake.

"Now, are there any questions?" Nobody raised their hands. "All right. Feel free to ask questions if you have any, but for now let's begin our study of science."

"We shall begin with the physical sciences, but before we do, it is important to introduce the basic mathematical concepts."

"I already know math," said Ash. "Is it okay to be excused from this class now?"

"Ah, not yet. It's very good that you know math, but these's so much more to the sciences than math. Since you know math, perhaps you can tell us what the four basic arithmetic operations are."

"Arith-what?" Ash exclaimed.

"Addition is when you combine two numbers together," said Joe.

"Subtraction is when you take one number away from another number," said Richie.

"Multiplication is when you add several groups of a number together," said A.J.

"Division is the inverse of multiplication, when you find how many times a number goes into another number," said Brock.

"Oh yeah, I know all of that," said Ash. "2+24, 4×416 and so on."

"Very good," said Professor Oak. "However, you're not always given an easy number statement like that. Very often, you'll be presented with a situation and you'll have to figure out by yourself what operation to use. For example, if Ash sends me five Pokémon, and Gary sends me seven, how many Pokémon would I have?"

"You'll have 12, because 5 plus 7 equals 12," said Tracey.

"That is correct. How about this? There are 20 Pokémon in a field, and then eight of them were caught by trainers. How many Pokémon would be left?"

"There will be 12 left, because 20 minus 8 equals 12," said Lara.

"That's very good. There are four members of the Elite Four, and each of them has five Pokémon. How many Pokémon must you defeat to defeat the Elite Four?"

"There will be 20 you have to defeat, because 4 multiplied by 5 is equivalent to 20," said Samurai.

"Correct. Finally, Ash sends me 180 Tauroses. Only 20 of them can fit in a box. How many boxes would I use up to store all of the Tauroses?"

"You'll need nine boxes, because 180 divided by 20 is 9," answered Misty.

"Yeah, I know all of that," said Ash. "Now can I PLEASE be excused from this class?"

"Us too," piped James.

"Now hold your horses," said Professor Oak. "There's much more to mathematics than this. For example, you'll have to learn how to work with negative numbers, fractions, and exponents."

"I know about negative numbers," said Duplica. "If you subtract a larger number from a smaller number, then you end up with a negative number."

"That's right," said Professor Oak. "Can anybody give me an example of this?"

"Yeah! Let's say I beat Ash in a Pokémon battle…" said Gary.

"That'll never happen!" Ash cried.

"Shut up! I beat Ash in a Pokémon battle, and he has to pay me 1,000, but let's say he only has 700. Well then, he has to pay me his 700, and he still owes me 300."

"That is very good," said Professor Oak. "In this case, 700 – 1,000 -300. You go past zero, and you have to keep going. Just imagine that there is a number line with 0 in the center. To the right of that is positive numbers, and to the left of that is negative numbers. As you add numbers, you go to the right. As you subtract numbers, you go to the left. This can create some interesting scenarios if you start out from the negative side.

"For example, if you have a negative number, like -50. If you add a positive number to it, then you go more to the right. However, the absolute value of the number would be smaller. Just think about it."

The students sat and thought about it. Misty spoke up. "So if you add 10 to -50, you'd end up with -40, because you went to the right ten times on the number line."

"Wait a minute," said Ash. "But 40 is less than 50! If you add a number, then shouldn't it be larger?"

"That's true," said Professor Oak, "but let's just take a negative number to mean owing money. Wouldn't it be better to owe only 40 than to owe 50? Therefore, -40 is actually greater than -50."

"Ah," said Richie, "on the other had, if you add a negative number to a negative number, then you'll end up being more negative."

"That's right," said Professor Oak, "but if you think about it, any time you have two signs that are different, you'd end up subtracting. For example, adding a negative number and subtracting a positive number are both equivalent to subtraction."

"So does that mean that if you have two signs that are the same, you'd end up adding?" asked Joe.

"That's exactly it. Even if you subtract a negative number, it's the same as adding. It might be difficult to think about it conceptually, but it is important to understand this.

"Multiplying and dividing with negative numbers follow the same general rules. If you multiply or divide two numbers with the different signs together, then you end up with a negative answer. However, if you multiply or divide two numbers with different signs, even if they are negative, then you end up with a positive answer."

"I get it," said Brock. "Like signs means positive, while unlike signs means negative."

"Correct," said Professor Oak.

"Next up we have fractions. Fractions may be a bit tricky to work with, but it will be important when it comes to the physical sciences. A fraction represents a part of a whole. Let's say that Misty made a pie.

"What type of pie?" asked Joe.

"I don't think that really matters," said Ash. "I don't think she can bake in the first place."

Misty grabbed Ash's ear and gave it a twist. "You may continue," she told Professor Oak.

"Okay. So Misty made a pie. We'll just say that it is apple pie. She cuts it into sixteen pieces. If she gives us each one piece, then each of us gets 1/16 of the pie. Since we all would have one piece out of sixteen pieces total. The 1 represents the part of the pie we have, and that is the numerator. The 16 represents the part of the pie that is in total. This is the denominator."

Gary spoke up. "So if I steal Ash's piece of the pie, then I'll have 2/16."

"And 2/16 is the same as 1/8, since you can divide two from both the numerator and the denominator," said Duplica. "That's what they mean by factoring."

"That's very true, Duplica," said Professor Oak, obviously impressed. "If you see that you can divide a number out of both the numerator and the denominator, then by all means do it. It will make your life much easier on the tests. For example, 2/4 is the same as 1/2. And 5/5 is the same as 1/1. If you have 1 as a denominator, then it is the same as the whole number. So 5/1 is the same as 5."

"And this is because you have five wholes?" asked Tracey.

"That is right. This idea would be useful when it comes to using fractions in arithmetic functions. When it comes to arithmetic in fractions, it is actually easier to multiply and divide. In multiplying, all you have to do is multiply the numerator and then multiply the denominator. So what would be 1/4 times 2/3?"

"The answer would be 2/12, which is the same as saying 1/6," answered Richie.

"Wait a minute, that doesn't make sense," cried Jessie. "1/6 is clearly less than 1/4. Why would you get a smaller answer when multiplying?"

"Let's just think of it this way. Let's suppose Misty made another pie. People have already eaten 3/4 of it, leaving the last 1/4 uneaten. Now you come over and see the remaining 1/4. You want as much as possible, but you don't want to eat the entire thing, so you take 2/3 from the remaining 1/4. So you end up with 1/6 of the original pie."

"I would have eaten the entire thing," said Jessie, "but I think I get it."

"So this would mean that multiplying fractions would get you a smaller answer, while dividing fractions would get you a larger answer?" asked Lara.

"That's not completely correct, but it's good to think of it that way. We'll get into the exception later. To divide fractions, you flip the fraction around so the numerator becomes the denominator and vice versa. Then you just multiply."

"So 1/4 divided by 2/3 would be 3/8," said Tracey.

"That is correct. 3/8 is larger than 1/4 because 1/4 is equivalent to 2/8," said Professor Oak. "Now, Lara, as far as the exception goes, what happens if you multiply 1/4 with 3/2?"

"Oh, I get it now. You would get a larger number when multiplying fraction if the fraction is larger than one," said Lara.

"That is correct. Adding and subtracting fractions is trickier. Before you can add or subtract, you'd have to make sure that the denominator is the same."

"Why would that be the case?" asked Samurai.

"Well, if you add or subtract a fraction, then you would always get a number that is larger or smaller, unless you're dealing with negative numbers. For example, if you want to add 1/2 and 1/3, you wouldn't add both the numerator and the denominator, because you would end up with 2/5, which is smaller than 1/2. Instead, you have to find the least common multiple of the two denominators."

"That would be six," said Todd.

"Correct, so the denominator would be 6. What would be the equivalent fractions of 1/2 and 1/3 if the denominator is six?"

"1/2 is equivalent to 3/6, while 1/3 is equivalent to 2/6," said Misty. "I figured that you have to multiply two by three to get six, so I have to do the same with the numerator. That's how I get 3/6."

"Very good. Once you have the same denominator, then all you can do is add the numerator."

"So 3/6 + 2/6 is 5/6," said A.J.

"That is correct. And what would be 1/2 minus 1/3?"

"1/6," Duplica answered.

"Correct. This is the basics about working with fractions," said Professor Oak. "The numbers would obviously be much more difficult, but the general concepts would be the same."

"With physics and chemistry, more often than not you'll be working with numbers that are very big or very small. For example, in chemistry, you'll learn about a concept of a mole. A mole is the same as saying a billion or a trillion, except that it's 602000000000000000000000, give or take a few quadrillion. When you're working with numbers like this, it's easiest to use something called scientific notation.

"Scientific notation is based on exponents. Exponents are basically a step after multiplication. It measures how many times you multiply a number by itself. So you all know that 2 x 2 is 4, and 2 x 2 x 2 is 8. An easier way to say 2 x 2 x 2 is 23. The three is the exponent, and it tells you how many times you would multiply a number. Multiplying by ten is a special case. Every time you multiply a number by ten, you just add a number by zero. So if you have 10 7, how many zeroes would there be?"

"There would be seven, since you basically multiplied ten seven times." said Richie.

"That is correct. Just think that there's a decimal to the right of a whole number. Every time you multiply a number by 10, you would move the decimal point to the right. It's the same when dividing a number by 10, except you would move the decimal point to the left. So 10 -7 would be .0000007. However, notice how this number only has six zeroes. This is because 10-1 is .1, since you only moved it to the left once."

"And 1 is 100, since you didn't move the decimal point at all," said Misty.

"Yes, Misty. This is very important, especially when it comes to multiplying or dividing exponents, but for now it's important to just understand how to get a number in scientific notation. To do so, you basically have to keep in mind that a very large number or a very small number is simply a manageable number multiplied or divided by an exponent of ten. So let's say you have a number like 3,469. That's the same thing as saying 3.469 x 1,000, or 3.469 x 10 3. Does anybody have any questions about this?" Professor Oak scanned the room, but nobody spoke up.

"Okay, I hope that everybody would be able to do these questions on the test. The reason we convert numbers into its scientific notation is so we won't be overwhelmed by numbers when we try to multiply or divide them."

Todd raised his hands. "What would you do if you want to add two numbers?"

"That's a very good question," said Professor Oak. "Adding or subtracting numbers with scientific notation is actually a tricky business. You really have to pay attention to the exponents. For example, if you have a number like 2.1 x 10 8 and 5.7 x 10 8, then you can just add the decimals before the number to get 2.1 + 5.7 is 7.8. However, if you have 2.1 x 10 8 and 5.7 x 10 7, it will not be as clear-cut. Does anybody know what you would get?"

Duplica answered. "The answer would actually be 2.67 x 10 8, because 5.7 x 10 7 is a decimal point smaller."

"That is correct. When you're adding and subtracting large or small numbers, it may actually be easier to add the numbers without scientific notation, since addition is the same no matter how many digits you have. With multiplication and division, it would be much more work the more digits you have, and scientific notation would be useful.

"The key thing to remember about multiplying numbers in scientific notation is that when you multiply them together, then you simply add the exponents. So 2.1 x 10 8 and 5.7 x 10 7 would get you 11.97 x 10 15, or 1.2 x 10 16. Similarly, if you divide these numbers, you would subtract the exponents. 2.1 x 10 8 and 5.7 x 10 7 would get you an answer of approximately .39 x 10 1, or just simply 3.9.

"Let's do a practice problem. Suppose Silph Co. manufactures over 9,000 Pokéballs every day, and they ship them all to 150 PokéMarts over the country. How many Pokéballs would each PokéMart get every day? Try to work this problem out by yourself."

The students scribbled furiously on pieces of paper, except for Ash, who looked befuddled, and Jessie, who was defiant. "Would somebody like to volunteer their answer?" Professor Oak asked when everybody finished.

"I got an answer of 60," said Misty.

"That is correct. Who else had an answer of 60?" All of the other students raised their hands except for James and the two that did not even try. "Would you like to do the problem on the board for the ones that did not get it?" asked Professor Oak.

Misty walked to the whiteboard and wrote down her steps, explaining them as she went on. "Well, this is a textbook example of a division problem, and one way to display a division problem is with fractions. 9,000 goes on the top, since that is your dividend, or the number you divide. 150 goes on the bottom, since that is the divisor, or the number you divide by. It's possible to do this problem with long division, but since we're learning how to use scientific notation, it's better to put 9 x 10 3 on the top and 1.5 x 10 2 on the bottom. Nine divided by 1.5 is six, and three minus two is one, so you end up with 6 x 10 1, which is the same as 60."

"That is very good, Misty," said Professor Oak. "And good job to everybody else who got the correct answer, even if you did not use scientific notation. Well, there are a few more mathematical concepts for us to cover, but it doesn't seem like we'll have the time to cover it today. If you've bought the textbook, be sure to read Chapter 1 of the Physics section. It will cover what we will go over next time so you'll be prepared. The topic will be basic Kinematics."

The bell rang, and the doors unlocked. Ash hurried to get out of the room, but noticed that Misty, Brock, and Richie were all waiting to speak with Professor Oak. As he waited, he shook his head and thought to himself, "This is going to be a long year."

--  
AUTHOR'S NOTE: I've decided to put an Author's Note for these chapters just to offer like a commentary about what I've written. Its probably pure selfishness on my part just because it's so bloody unnecessary. Anyways, this is the first chapter. I wanted to have an opening where we introduce all of the characters, not that they'd play much of a big role. In the story that I originally wrote two years ago that served as like a preview, there were only four students: Ash, Misty, Brock, and Gary. When I started this, I knew that I wanted to have more than just four students, so I racked my brain and found eight more people that were around Ash's age to be students. And then I decided that Team Rocket were important enough to be students enrolled against their will. I wanted the roll call in the beginning to be like an actual classroom, where the teacher would call out the students' first and last names. There was on problem: most of them didn't have last names. I had to make them up. Slate and Waterflower are like the unoffcial last names for Misty and Brock in the online world, so I just used them. For Duplica, I took the "House of Imite" thing and applied it here. A.J. was a Sandshrew trainer, and Sandshrew is a Ground Pokemon. Ground is kind of a stupid nickname, so I went with Stone, because they're both similar in that they're weak against water. For Samurai, I took the most obvious Japanese last name among Nintendo fans: Miyamoto. Joe, who if you don't remember was the kid the trio meet in the episode "The School of Hard Knocks" got the last name Smith just because Joe Smith is such a great name. And Richie? I just grabbed it out of nowhere. As for the actual writing, I've never been very good at prose, but I never knew how bad until I started writing this story and have it be dialogue upon dialogue. I suppose it's forgiveable in this case since it's not a very serious story anyways, but I'd definitely have to fix this if I ever want to develop my writing. As far as the science goes, I decided to start off with the basic mathematical concepts that would be useful in the MCAT, since the students would probably have to do it for real. It was pretty easy come up with interesting examples in this chapter and some of the other early chapters, but unfortunately things won't be like that.


	2. Chapter 2: Kinematics

Brrring

DISCLAIMER: Once again, I don't own Pokémon or the scientific material being discussed in this story. Nintendo and Shogakukuan own Pokémon...at least Shogakukuan owns the Pokémon anime that this story is largely based off of, and the science is open for everybody to use and learn. So use and learn it. Special thanks go to the Kaplan program for the books that I've used for these chapters, and also my Kaplan instructor Matt Kinney for his diligence in proofreading the chapter for scientific accuracy. In addition, I'd like to thank Angelo Godbey for showing me I still don't have an idea of what I'm talking about when it comes to kinematics in an x and y direction. You use the vf vo + at equation to find the time it takes for the projectile to stop, multiply it by two to get the time it takes to get back to the origin only in a different direction, and the y yo + vot + ½ at 2 equation to find the time if the projectile goes down beneath the origin.

--

Ash, Misty, and Brock entered the classroom before the bell rang for their second day of classes. "Another day of toil and trouble," muttered Ash.

"Don't be like that," admonished Misty. "Some of this stuff is pretty interesting."

"And by the way," asked Brock, "did you finish reading the chapter, Ash?"

"How can I? I fell asleep about five minutes into it. The book was SO boring!"

"Well, I must admit it was a bit on the didactic side," said Misty, "but there are far worst ways you could use to teach this material."

Richie ran up to greet his friend. "Hey, Ash, how did studying go last night?"

"Don't even talk about it," Ash groused. "Did you manage to finish the chapter?"

"Yeah, I finished the first chapter and even went on to the second. If this material is proving to be too difficult, I can always go to your place and help you study."

"That would be great," said Ash. "If only we would be able to meet up without this stupid science stuff hanging over our head."

"Even if you do go and help him, you'd just be wasting your time," said Gary.

"Why you…Gary! I'll show you!"

"Oh yeah? Let's see if you can answer this easy question. If I throw a Pokéball 60° degrees in the air with a velocity of four meters per second, and it goes four seconds before it hits the ground, how far away from me is the Pokémon that I caught so gracefully?" Gary's cheerleader cheered behind him.

"Uh, I don't think it's physically possible to have a hang time of four seconds if you only throw the Pokéball four meters per second," said Brock.

"Ah, it doesn't matter. The only thing that matters is whether or not Ash the loser here has the ability to solve this problem."

"I don't think you should try," said Richie, "especially if you didn't read the first chapter."

"Who cares?" cried Ash. "Any idiot would know that it would go sixteen meters, because 4 x 4 is 16!"

"Hah! That's incorrect! I knew you were too stupid to get it!" Gary laughed triumphantly.

"What? How is that possible?" wailed Ash.

"Actually, he's right. The correct answer would only be eight meters," said Misty.

"How can that be?"

"If you read the chapter, you would have known that when the velocity is at an angle, you'd have to consider the horizontal and the vertical components separately."

"That's right, loser!" Gary sneered. "How is it that you have such smart friends and still be so stupid?"

"Don't call Ash stupid, and he's not a loser," Richie cried.

At that moment, the bell rang and Professor Oak walked into the classroom, late but still composed. "All right class," he said, "settle down. We have a lot to cover today."

Team Rocket also scrambled into the room, beat up, dusty, and out of breath.

"What happened to you guys?" asked Tracey.

"We went to the twerp's house to get Pikachu," Jessie grumbled, "and the stupid Mr. Mime his mother has chased us away."

"At least Mr. Mimie can be useful at times," Ash said dryly.

"Ah, so you were doing a bit of running in the morning?" asked Professor Oak. "Perhaps we would be able to use your example in our study this morning."

"Uh, we'd rather not," said James.

"Come now, don't be bashful. Anyways, I hope everybody took the time to read over the first chapter, because this would serve as the foundation for much of what we are going to do in physics: Kinematics. Would anybody like to tell us what the three basic quantities that will be useful in kinematics?"

Todd raised his hand. "The first quantity is displacement. This measures how much you have travelled. The basic units for displacement would be meters. The next quantity is velocity, which measures your change in displacement over a change in time. The basic units for velocity would be meters per second. The final quantity is acceleration. This is the change in velocity over the change in time. The unit for acceleration is meters per seconds squared, since it is change in velocity (m/s) per second."

"Very good," said Professor Oak. "I'm glad that you did your reading. Todd is correct. Displacement, velocity, and acceleration are the three important quantities in kinematics."

A.J. raised his hand. "Yeah, I was reading the chapter, and it sure seemed like displacement is no different from distance, and velocity is no different from speed, yet they apparently aren't the same thing at all. It really messed me up when I did the questions at the end of the chapter. What is this difference they were talking about anyways?"

"Ah, great question, A.J. Distance and speed are very closely related with displacement and velocity, but there is one difference that separates them. Can anybody explain what this difference is?"

Misty spoke up. "Distance and speed are scalar quantities, meaning that they are independent of direction. If you walk from Viridian City to Pewter City, then from Pewter City to Cerulean City, then the distance you walked is the total distance from Viridian to Pewter, and then from Pewter to Cerulean. However, your displacement is the straight-line difference between Viridian and Pewter. This is because displacement and velocity are vector quantities. Every time you change a direction, even if your speed doesn't change, your velocity would change."

"That is exactly it," said Professor Oak. "This is a good transition into what we were going to cover last class but did not have a chance to do so: the concept of vector addition and breaking a vector down into its components.

"As Misty said, vector quantities are dependent on your direction. If you go north and then go east, then your displacement and velocity would change. It is helpful to draw vectors as lines with an arrow pointing in the desired direction: ––––––––––––

"Whenever you go in one direction for a long time, then the vector quantities are no different from scalar quantities. However, whenever you change directions, the vector quantities would change, and you would have to mathematically determine what the new quantity would be. For example, let's just take Misty's example of walking from Viridian to Pewter, and then to Cerulean. Suppose the distance from Viridian to Pewter is 30,000 meters, while the distance from Pewter to Cerulean is 40,000 meters. Can anybody figure out what your overall displacement is?"

Duplica instantly raised her hand. "Well, 30,000 meters is 3 x 104 while 40,000 meters is 4 x 104. You would draw a straight line that's going north, and another straight line going east. The first line would be labeled 3 x 104 while the second line would be labeled 4 x 10 4. The overall displacement would be the straight line from where the first line begins to where the second line ends. You would use Pythagorean's Theorem, or X 2 + Y 2 is Z 2 to figure out the length. X 2 and Y 2 would be 9 x 10 8 and 16 x 10 8 respectively. Adding those two up would give you 25 x 108, and the square root of that would be 5 x 10 4, so your final answer would be 50,000 meters."

"Very good, Duplica," said Professor Oak. "Now, would you know what the final direction of the displacement would be?" Duplica quietly shook her head.

"It's okay, because this would require a general knowledge of trigonometry, or the study of triangles. There are three trigonometric functions that are dependent only on a particular angle in a right triangle. Here is a sample triangle XYZ, where the angle YXZ is 60° and angle XZY is 30°. We'll consider angle XZY, which we have already established is 30°.

"The first trigonometric function is sine, which is essentially the ratio between the lengths of the opposite side XY with the hypotenuse XZ. No matter how long each of these sides may be, the ratio of the length would be the same if the angle is the same. The cosine of the angle is the adjacent side YZ with the hypotenuse XZ. Similarly, the cosine of an angle remains the same no matter how much you expand or decrease the triangle. Finally, the tangent is the opposite side XY over the adjacent side YZ.

"If I tell you that the sine of 30° is ½, and XZ is 6 meters long, then how long would XY be?"

Meowth raised his hand. "Oh yeah, I know this one. XY would be 3 meters long."

"Very good," said Professor Oak as Jessie whacked Meowth on the head for the audacity of actually participating in class. "Now, let's take this right triangle Duplica described. We'll call it VPC, for Viridian Pewter Cerulean. On one straight edge is 30,000 meters, and on another straight edge you have 40,000 meters. The hypotenuse is 50,000 meters. The angle we are concerned with is this angle PVC, because by figuring out this angle, you would be able to figure out the direction of the vector. Does anybody want to take a crack at solving this?"

Brock took a crack. "Well, we know the length of all of the sides, so we can use any one of the trigonometric functions. I'll use sine, since we've done so much with it. Sine is opposite over hypotenuse, so it would be 40,000 meters over 50,000 meters, which is .8. Our answer would be the angle with a sine of .8."

Tracey took out his calculator to figure out the answer. "The angle with a sine of .8 is 53.13°, so the direction of our displacement would be 53.13°."

"That is correct," said Professor Oak. "These trigonometric functions would be very useful when it comes to adding or subtracting vectors, which is what you may have to do. Let's suppose you have two vectors 2 meters long, one making a 30° angle to the horizontal, and the other making a 60° angle to the horizontal. If you want to add them together, you would put them end to end. However, this is not enough. In order to find the magnitude and the direction, what you have to do is to find the _x_, or horizontal, and _y_, or vertical components.

"Let's just take this first vector. It should be easy to find the _x _and _y_ components if you make this into a right triangle. You just draw two lines going up or down from the edges of the vector until they meet. Voila! You've figured out the identity of the components, and now all you have to do is use the trigonometric functions to find the magnitudes. Tracey, can you use your calculator to tell us what the _x_ and _y_ components are?"

Tracey worked it out. "The _y_ component would be sine of 30° times two, and that would be 1 meter. The _x_ component would be cosine of 30° times two, and that would be 1.73. For the other vector, the _y_ component would be sine of 60° times two, and that would be 1.73. The _x_ component would be cosine of 60° times two, and that would be 1."

Misty did the math. "The sum of the _x_ components would be 2.73, and the sum of the _y_ components would be 2.73. However, these are only the components, and the final vector would be the hypotenuse of the resulting right triangle."

Tracey plugged numbers into his calculator and came up with an answer. "The final sum would be divided by 3.86 is .707, and so the direction would be 45°."

"That would make sense," said Professor Oak. "The two sides both have the same lengths, so this would be an isosceles right triangle, and the angles in one of those is 90°, 45°, and 45°.

"Most of the sines and cosines you can find with a calculator, but there are a few that you would need to memorize. For example, I already told you that the sine of 30° is ½. The cosine of 30° is √(3) / 2. Similarly, the sine of 60° is √(3) / 2, and the cosine of 60° is ½. The sine and the cosine of 45° are the same, and they are both √(2) / 2. The sine of 0° is zero, and the cosine of 0° is 1. The sine of 90° is 1, and the cosine of 90° is 0. Those are the sines and cosines that you should memorize.

"The last thing you should know about vectors is that it is possible to subtract vectors. To do this, you would just have to flip a vector over, and then add the vectors normally.

"With the introduction to vectors and trigonometric functions out of the way, we can now finally focus on the topic at hand. As Todd said, the important quantities in kinematics are displacement, velocity, and acceleration. All three of these quantities are tied together in some way. Now, Jessie, you said that you were chased away by Mr. Mime. How fast would you say you were going when you started running and at your top speed?"

Jessie didn't say anything, but James said, "Well, I don't think we were moving at all when we started running, but by the time we reached our maximum speed, we were going probably two meters per second."

"And how long did it take you to get to that top speed?"

"Uh, I would guess ten seconds," said James while Jessie made efforts to get him to keep his mouth shut."

"Okay, thank you, James. If we assume constant acceleration, we have enough information to calculate the average velocity, acceleration, and even distance traveled. So, who would want to take a shot at calculating the acceleration?"

"I would," said Lara. "They went from zero m/s to two m/s, so that would be their change in velocity, or delta v. The time it took them to do so is ten seconds. The equation for acceleration is change in velocity over change in time, so it would be 2 m/s / 10 s, and that would be .2 m/s 2."

"That is correct," said Professor Oak. "It is possible to interconvert between displacement, velocity, and acceleration using the equations Todd said at the beginning of class. Now, we know that acceleration usually always changes, but in many of the problems we do on paper, we have to assume constant acceleration. If we make this assumption, then there are four equations that you can use to find displacement, velocity, acceleration, and time even when one of those quantities is missing. Who can tell me what the four equations are?"

"The first equation is vf vi + acceleration change in time," said Gary. "You could use this to find delta v, acceleration, and change in time even if you don't know the displacement."

"The second equation is the square of the initial velocity plus two times the acceleration multiplied the change in distance is equivalent to the square of the final velocity," said Samurai. "You may use this equation to find the change in velocity, the acceleration, or the change in displacement even though you do not know the change in time."

"The third equation is △d vit + ½ at 2," said Richie. "You can use this to find the change in displacement, the initial velocity, the acceleration, or the time even if you do not know the final velocity."

"These are all correct," said Professor Oak. "And what is the final equation?"

Duplica spoke up. "The final equation is △d ½ (vi + vf) x △t. This equation would allow you to find the displacement, the change in velocity, or the change in time even if you do not know the acceleration."

"That's very good. These are the basic equations you would have to remember for one-dimensional kinematics."

Ash was aghast. "What? Do you mean there's two-dimensional kinematics?"

"That is correct," said Professor Oak, "but before we go into two-dimensional kinematics, we'd have to discuss a special case of one-dimensional kinematics: free fall.

"I am sure you all remember Team Rocket's spirited entrance yesterday. That is a perfect example of free fall, since they were falling with a constant acceleration."

"The acceleration due to gravity," said Misty.

"That is correct. The Earth has gravity, and it has the same pull on anybody no matter where on Earth you are, as long as you're still on the surface and not miles underground in a Diglett's tunnel, that is. The acceleration g is equal to -9.8 m/s2, although it is sometimes acceptable to round that to 10 m/s2. Using these kinematic equations, we can calculate anything about the free fall, from the final velocity to the time it takes for something to fall. There is one thing you must remember, and that is that the acceleration due to gravity is often negative when you use it in the equations."

"Why is that?" asked A.J.

"As we said, it is because these vector quantities are dependent on direction. If we are doing one-dimensional kinematics, it is important to signify one direction as positive, and so the other direction would be negative. Take acceleration for example. When James was speeding up to get away from the Mr. Mime this morning, we would say that his acceleration is positive. Once he gets far enough away from it, he would slow down."

"And when he slows down, his acceleration would be negative," said Brock.

"That is right. In addition, you can do it for velocity and displacement as well. Let's suppose Ash is going north, from Viridian City to Pewter City. After that, he goes from Pewter City to Viridian City. What would his net displacement be?"

"The net displacement would actually be zero, since he went back to where he started," said Lara.

"Ah, that's because we're assuming the northern direction to have a positive displacement, and the southern direction to have a negative displacement," said A.J. "The two would cancel each other out of they're opposite but equal. I get it now."

"That's terrific," said Professor Oak. "It's the same if you're doing a free fall problem. Our positive for acceleration is always in the upwards direction, which doesn't make any sense because it is physically impossible to toss something like a Pokéball in the air and have it accelerate upwards. A Pokéball would slow down and eventually falls if we throw it up in the air. This is our negative acceleration.

"So let's think back to when Team Rocket fell from the ceiling yesterday. We'll say that it is 3 meters from the floor to the ceiling, and the group is starting from rest. How fast would they be going when they hit the floor, and how long would it take them?"

With the kinematic equations in hand, the class began tackling this deceptively simple problem. Misty was the first to put her pencil down, but she waited until everybody was finished before saying her answers.

"Well, we know the initial velocity, the displacement, and the acceleration. We can find the final velocity and the time with the vf 2 is vi 2 + 2a△d and the △ d vit + ½at 2 equations respectively. In this situation, it is essential that we assume the negative direction to be positive, because the initial velocity is zero, and the quantity we want to calculate is a square."

"Ah, that's a very good point," said Professor Oak. "While it is customary to use the upwards direction as the positive, it may sometimes be essential to make the downwards direction positive, mostly if you're starting from rest. I hope most of you caught this." Most of the students nodded, but a few of the students were abashed. "Please continue."

"Okay," said Misty, "for the first equation, the initial velocity is zero, the acceleration is 9.8 m/s2, and the displacement is 3 meters. Plugging all these into the equation, it would be 2 x 9.8 x 3, which is 58.8 m2/s2. Taking the square root of that would tell you that their final velocity was 7.68 m/s. Now that we know final velocity, we can actually use any equation, but using the △ d vit + ½at2 equation, we rearrange it to find out that 2△ d / a t 2. Plugging everything in, we get that t 2 is around .6, so the square root of that is around .78 seconds."

"It sure seemed longer than that as we were falling," said James.

"That is correct," said Professor Oak. This is the basics of free fall. We can use the same equations if we consider a situation where we throw an object in the air, except this time the initial velocity would be something other than zero, and we cannot get away with using a positive acceleration. This throw can be straight up in the air, or to the side. In the latter situation, the kinematics is known as projectile motion.

"Gary already touched on the important thing about projectile motion when he quizzed Ash before class. The important thing to know is that you must take the _x_ and the _y_ components independently. The _x_ component is the easy one, because the horizontal acceleration of a projectile is zero if we assume no external forces like air resistance – friction by the air – is in play. This is related to Newton's First Law of Motion, which we will discuss in the next class. The _y_ component is just like what we talked about with free fall. The only thing to keep in mind is that at the top of the projectile's parabolic path, the velocity is zero. So Gary, can you repeat what you told Ash before class?"

"Of course I can," Gary said confidently. "If I throw a Pokéball 60° degrees in the air with a velocity of four meters per second, and it goes four seconds before it hits the ground, how far away from me is the Pokémon that I caught so gracefully?"

"Thank you, Gary. As Brock said, the time Gary said can be discounted in this case, but it's understandable why Gary would mention it because he was testing Ash about the fact that horizontal motion has zero acceleration. However, we would have a new time and a new horizontal displacement if we want to make this Pokémon capture physically acceptable. Who would be willing to take a shot at figuring out the new location of Gary's new Pokémon."

"I'll do it," said Richie. "We'll make the assumption that the Pokémon is at the same height as when Gary releases the ball. A projectile takes a parabolic motion that is symmetrical as long as the initial height is the same as the final height. To find the time, all we would have to do is to find the time when the Pokéball is at the apex of its path, and double that. At the peak, the Pokéball has a vertical velocity of zero, so we can use the equation vf vi + a△t to find the time. The final velocity is zero, so we'd be able to tell that the initial velocity is equal and opposite of the acceleration times the change in time. If we rearrange the equation so that the time is on one side, we get △t vi / a. We would need the overall velocity times the sine of 60°, and that is 3.46 m/s. Divide that by the acceleration 9.8, and we get .35 seconds. Double that and you get .70 seconds.

"To find the horizontal displacement, you would just multiply the horizontal velocity, four times the cosine of 60°, by the time. This is the equation △x 1/2(vf + vi) △t, only the initial and final velocity are the same. The cosine of 60° is ½, so the horizontal velocity is 2 m/s. The Pokémon would only be 1.4 meters away."

"Hah!" cried Ash. "That's only like the distance from me to Brock! That's not very impressive at all!"

"Yeah? Well I'll bet you'd miss the Pokémon even when it's that close," Gary countered.

"All right, that's enough from you two," said Professor Oak. "That is correct, Richie. If you want to figure out the time if the Pokémon is lower than the point of release, then you would just use the △d vit + ½ a t2 equation with the negative initial velocity as the starting velocity and add the time to the .7 seconds you got. If you want to figure out the time if the Pokémon is higher than the point of release, then you would use that equation with zero as your initial velocity and add it to the .35 seconds you got at the top.

"Well, our time is almost up. Your homework for tonight is to do the homework problems for Chapter 1 if you haven't done so already, and also to read Chapter 2. In addition, I'll throw in an extra credit problem: what was the maximum height of Gary's Pokéball? See you all tomorrow."

The bell rang, and the students slowly streamed out of the room.

"Oh, that Gary," Ash muttered. "I'll get the best of him, if it's the last thing I do!"

"Well, if it's in physics, I don't think that'll ever happen," said Misty.

--  
AUTHOR'S NOTE: I think the personalities for each of the characters are coming out with this chapter. There are the people that will obviously be good students, people that would obviously be bad students, and people that just don't give a darn (Team Rocket.) Then there are the people like Joe who fade away into obscurity. I think I completely forgot about him in a few of the early chapters, but how could you not be anonymous with a name like Joe Smith. And then there is the rivalry between Ash and Gary, which is the primary plot point outside of all of the science junk. Gary's not too bad in these early chapters, but I've definitely exaggerated his pompous behavior and descending attitude towards Ash and everybody else. You'll see what I mean.

Anyways, as I said in the disclaimer, I made an egregious error in the calculations of Gary's Pokéball example near the end of the chapter. I think I used the y yo + vot + ½ at 2 equation and set y equal to zero. Thankfully I checked it over with somebody first. The rest should be okay. The only major integration of Pokémon I could think of in this chapter is the Pokéball example and the three major Pokémon cities in the discussion about vector addition. The lack of superscripts, subscripts, and equal signs are annoying, just because so many equations have these values. Why, ? WHY?


	3. Chapter 3: Newtonian Mechanics

Brrring

DISCLAIMER: I don't know if I have to do this for every single chapter, but better safe than sorry. Plus, it increases the number of words I have. Nintendo owns Pokémon, and Shogakukuan owns the rights to Pokémon anime which includes the anime Misty that I'm so hopelessly in love with. The material covered in this story was based on a Kaplan lesson book, and thanks to Matt Kinney my Kaplan instructor for his tireless efforts of reading this.

--

It was the beginning of another school day. Ash, Misty, and Brock walked into the classroom, where Richie greeted each of them with a genial hello.

"Hi, Richie," said Ash. "Thanks for taking the time to help go over the stuff for Chapter 1 with me yesterday."

"No problem," Richie replied. "It was the least I can do. So, did you try that extra credit problem Professor Oak posed?"

"Uh, no, I didn't."

"I did," said Brock. "What did you get as an answer?"

"I got approximately .6 meters," Richie replied.

"That's what I got too! What equation did you use?"

"Well, I used the △y is ½ (vf + vi) △t equation."

"Oh, well I used the vf 2 is vi 2 + 2a△d equation."

"Well, you should get a similar answer no matter what equation you used," said Misty. "Richie's equation is easier to use, since you only need the vi of 4√3 / 2 m/s and the △t of .35 seconds. Multiply those together and you get .61 meters. For Brock's equation, you know that the final velocity is 0, so you can move the –2a△d expression on the other side, and solve for △d that way."

"I take it you did it both ways," said Richie.

"That's right. It was a way to check my work."

The bell rang and Professor Oak walked into the classroom. Once again, Team Rocket came hurrying in before Professor Oak shut the door.

"Don't tell me you were trying to sneak into my house again," cried Ash.

"Actually, we were waiting for Jessie to put on her makeup," James replied.

"Okay, everybody, it's time to settle down," said Professor Oak. "We have a lot to cover today, but first, I hope all of you tried the extra credit." Most of the class raised their hands. "That's very good. If you've done it, please pass your work into me." Once all of the papers were in, Professor Oak announced, "Now that none of you can change your work, I can safely tell you that the answer was .61 meters." Some of the students celebrated silently while others seemed crestfallen.

Professor Oak waited until everybody was still before continuing. "Today, we will be talking about the work of Isaac Newton."

"Who is he? A famous Pokémon trainer?" asked Ash.

"Actually, he is one of the greatest scientists of all time," said Professor Oak. "He came up with much of the concepts that make up modern physics, and we'll be studying some of his most famous contributions today: his laws of motion and his universal law of gravitation. With this, we will also introduce the important vector concept of force. A force is anything that acts on an object or a particle, and we'll be using it throughout our study of physics. What are some common forces that you can think of?"

Duplica raised her hand. "Your weight is the force due to gravity. It's equal to your mass times the acceleration due to gravity, which we established last time as -9.8 m/s 2."

"Yes, that is the most basic force. You'll be using it in almost every situation. It also important to keep in mind of the definition of weight as Duplica said it. Are there any other forces?"

"Friction is a force that opposes the force of movement," said A.J.

"Yes, although there are two different frictional forces. We'll be talking about this later in the class," said Professor Oak.

"Tension is the force felt in a rope that is connected to an item," said Lara.

"Air resistance is a type of force that slows down something that is falling. It's like friction by air resistance." Brock answered.

"The normal force is the contact force that an item exerts on a surface that it rests on," stated the Samurai.

"These are all very good answers," said Professor Oak. "Now, Newton's Laws of Motion relates forces to an object's movements. His first law is known as the Law of Inertia. Let's suppose I have a Pokéball, and I give it a push on top of a frictionless surface. What do you suppose will happen to it?"

"I know," cried Ash. "It's going to eventually stop!" Feeling proud of himself, Ash looked around and saw Misty shaking her head and Gary snickering to himself. "What's the matter?" he asked.

"You idiot," Gary said gleefully, "there's a reason why gramps made note of the word 'frictionless.' An object in motion would stay in motion unless an external force acts on it. That's the law of inertia."

"That is correct, Gary," said Professor Oak. "The law of inertia states that an object at rest will stay at rest, and an object in motion will stay in motion unless a net force acts on it. Friction is a force, which is why most things would eventually come to a stop.

"The second law of motion is the basic definition of a force, and it is one of the most important concepts you will learn. The law states that the net force is equal to the mass of an object times the object acceleration. You can play around with the equation to state that an object's acceleration is equal to the net force divided by its mass, or that an object's mass is equal to the net force divided by its acceleration, but the equation stays the same either way. You can also use the equation to figure out the unit of force: the Newton, or kgm/s2.

"It is important to know that like acceleration, force is also a vector, so different forces would be able to act on the same object, which explains why it is only the net force that is included in this equation. This would explain the law of inertia. The Pokéball that sits there on our frictionless surface has no acceleration because the net force is zero. If we give it a push, then that is applying a net force to it, causing it to accelerate. After that, however, the Pokéball keeps going at the same speed because its acceleration falls back down to zero."

"What about the object's weight? Isn't that a force on the Pokéball?" asked Todd.

"Yes, there is the force of gravity on the Pokéball, but there is another force in the equal and opposite direction. Does anybody know what it is?"

"It is the normal force," said Samurai. "The frictionless surface would be applying a constant, upward force to the Pokéball that cancels out the weight."

"Very good," said Professor Oak. "Mass is the only quantity in the equation that is scalar. It measures an object's inertia and has a unit of kilograms. This means that something massive like Snorlax may weigh less in outer space, where the acceleration due to gravity is something very close to 0 m/s2, but it still requires a lot of force to get it going.

"The third law of motion states that every force is opposed by an equal and opposite reaction force. For example, if you have a rocket…"

At that moment, Jessie jumped up and cried, "Team Rocket blasts off at the speed of light!"

"Surrender now, or prepare to fight!" James shouted.

"Meowth! That's right!"

After a long and uncomfortable silence, Professor Oak continued. "It's good that you three are so emotional about rockets, because a rocket accelerating in space requires Newton's third law of motion. A rocket accelerates by spewing particles out at a certain speed. The force that the rocket spews out the particles is the same force that the rocket would experience, and as the Fma equation shows, this would affect the acceleration.

"It is important to note that weight and the normal force does not qualify as a pair in Newton's third law. This is because forces in Newton's third law apply to different objects. The weight and the normal force on our Pokéball are both acting on the Pokéball, so that would disqualify it from Newton's third law. Does anybody know what the equal and opposite force to the Pokéball's weight is?"

"The equal and opposite force is the Pokéball's force of gravity on the Earth," said Misty.

"That is correct," said Professor Oak. "Everything has an attractive force of gravity on something else. For example, Misty, did you know that there is a force of gravity between you and Ash?" After hearing this, Misty and Ash quickly scooted their chairs away from each other.

"The force of gravity is relatively small compared to some of the other major forces. The only reason why we can feel the gravity of the Earth is because the Earth is so massive. Newton was able to come up with the formula of the attractive of gravity: F Gm1m2 / r 2, where r is the distance between the centers of gravity of two objects, and G is constant with the value of 6.67 x 10-11 N m 2 / kg 2. Now, suppose Ash is 40 kilograms and Misty is 45 kilograms, how close would they have to be for the attractive force of gravity to become 1 N?"

Tracey did the math on his calculator. "If we rearrange the equation, we find that the distance is equal to the square root of G m1m2 / F. The product of Ash and Misty's masses would be 1.8 x 103 kg 2, and you multiply that with G to get 12 x 10-8 Nm 2. The force on the bottom leaves the magnitude the same while cancelling out the Newtons. Take the square root to leave 3.46 x 10 -4 m."

"Yes, this means that the centers of gravity for both of them must be .3 millimeters apart to feel even one Newton of force.

"Our acceleration due to gravity of 9.8 m/s 2 is derived from this equation. While calculating the force of gravity we have with the Earth, m1 would be the mass of the Earth, and r would be the distance from the surface of the Earth to the center. It would be a major pain to enter these values every single time, but if you multiply them all together, you get the acceleration of 9.8 m/s 2. It is still important to keep in mind the main force of gravity. The further you get from the surface of the Earth, the larger r gets, and the smaller the force due to gravity would be.

"The other important force is friction. As A.J. said, frictional force goes against the directional force. There are two types of frictional forces, and they are both related to the normal force of an object. The first type is static friction. This is the friction that keeps an object like this table or a Snorlax in place even if you apply force to it. The static friction is actually a range of value. There is a maximum value, and once the maximum value of static friction is reached, then the object would start moving. To calculate this maximum value, you need to take the coefficient of static friction, which is different for each object, and multiply it with the normal force on the object.

"The kinetic friction is a set, speed-independent value. It is the friction that opposes you as you try to move an object. To calculate the kinetic friction, you would need to multiply the normal force with the coefficient of kinetic friction.

"So, suppose you have a Snorlax, who has a mass of 460 kg, and let's just say that its coefficient of static friction is .8 and its coefficient of kinetic friction is .5. What force would you need to put in to get it to start moving, and what force would you need to keep it accelerating?"

Richie answered. "If we assume that the Snorlax is on level ground, then its normal force would be about 4600 N. You would need to put in 3680 N of force to overcome the static friction and have it start moving, and you would need 2300 N of force to keep it accelerating."

"Very good," said Professor Oak. "Always keep in mind that the coefficient of kinetic friction is almost always less than the coefficient of static friction. Richie also made a good assumption that the Snorlax is on level ground? What would you think would happen if it was on a hill with an incline of 30°?"

"The frictional force would be less, of course," said Todd.

"Correct, and why would that be?"

"That is because the frictional force depends on the normal force," said Samurai. "The normal force is always perpendicular to the surface that the object lies on. If an object is on an incline 30°, then the normal force would be 30° to the vertical axis."

"Yes, and the weight of an object is always pointed downwards," said Duplica. "This way, if you want to find the normal force of an object, you have to break the weight down into its _x_ and _y_ components. The normal force would be equivalent to the _y_ component of gravity, so to find this _y_ component, you would need to multiply the weight with the cosine of 30°, so the normal force of the Snorlax on the incline would be 3984 N."

"And the _x_ component of weight is the force of gravity in the _x_ direction," said Brock. "If we are pushing the Snorlax at a constant acceleration, then this value would be equal and opposite of the kinetic friction. However, the maximum value of the Snorlax's static friction would be about 3172 N, and the force of gravity along the _x_ direction is only 2300 N, so the Snorlax would just sit there and sleep."

"Yes, that is a very complete picture of this problem," said Professor Oak. "Not only did you figure out the forces of friction, but you also painted of picture of translation equilibrium and what a free-body diagram would be like. A free-body diagram is very important in figuring out what all of the forces of an object would be, and you may have to break a vector down into its _x_ and _y_ components if it is in an angle. For our Snorlax, since it is resting on an inclined plane, the weight is now 30° away from the _xy_ axis, which is why as Duplica said we must take the _x_ and the _y_ components. Once we figure that out, we can easily see that the normal force opposes the _y_ component of weight, and the static friction is beneath its maximum value and also opposes the _x_ component of weight. Since the Snorlax would not move, it is in translational equilibrium.

"Whenever you want to do a problem related to translational equilibrium, it is important that you draw free-body diagrams, and that the net force in the _x_ and the _y_ directions are equal to zero.

"There is another form of equilibrium, and it is rotational equilibrium. As the title states, this form of equilibrium is related to the angular motion of an object, which means that if a system does not rotate, it is in rotational equilibrium. Rotation equilibrium requires that torque is equal to zero. The torque is the angular force of an object, and it is equal to the distance of the force from the center of the rotating item multiplied by the component of the force that is perpendicular to the item. Since we usually give the angle of force as the difference from the _x_ axis, then we would use the sine of the angle to figure out the component.

"You can easily see torque at work when it comes to opening and closing a door. If you press on the handle of the door right where it connects to the door, then you can push as hard as you want, hard enough to get a Snorlax to move, and nothing would happen, because the distance between the force and the center is equal to zero. Similarly, if you push against the handle, then you still won't get anywhere, because the angle is 0°, and the sine of 0° is zero. If you want to open a door, then it's best to push directly down on the handle at the edge of the handle to maximize the distance and the sine of the angle of force."

"Oh, so that's how you open a door," James said happily. "I'll have to try that the next time I see a door handle."

"There is one last thing we will cover in this class, and that is the idea of centripetal force." Professor Oak tied a Pokéball to a string and started swinging it around. "If you tie a Pokéball to a string and spin it around, it may go in the same velocity, but the Pokéball is accelerating the entire time."

"That's because velocity is a vector, and it's dependent on direction. Since the direction is constantly changing, the ball is accelerating," said Lara.

"Very good," said Professor Oak. "This acceleration is known as centripetal force, and it is directed towards the center of the spin. The value of the acceleration is v 2/r, where r is the distance between the object spinning and the center. In this case, the r would be equal to the length of the rope. The centripetal force is in the same direction of the acceleration, and is equal to mv 2/r."

At that moment, the string broke. It flew and hit Ash square in the face. "Oh, I'm sorry, Ash, but this shows you that at any moment, the Pokéball's velocity is in a direction perpendicular to the force, which is directed to the center of the string. The Pokéball would continue to go that way, but the force would cause it to change direction. This is how satellites are able to stay in the air. The force of gravity is the centripetal force."

The bell rang, and Ash rushed out of the room to get his bloody nose taken care of, making sure to get the maximum torque while turning the door handle. Misty and Brock looked at each other and shook their heads. "It's going to be a long year," said Brock.

--

AUTHOR'S NOTES: This chapter was a little bit on the short side, especially compared to the ultra-epic first two chapters, but it had some clever Pokémon examples, even if most of them involve lifting a Snorlax. But then again, what better way is there to talk about forces without using the Snorlax? I thought the incident where Team Rocket screamed their motto at the mention of a rocket was a bit lame since the joke was blah and it was such a disruption, but I left it in because I'm taking this as a real classroom, and that's probably something that Team Rocket would do.


	4. Chapter 4: Energy and Momentum

Brrring

DISCLAIMER: Nintendo owns Pokémon. They make billions off of it, which begs the question why we still shamelessly consume all the Pokémon stuff we find. Then again, I'm writing this while staring at my Pokémon Indigo League DVDs, but VIZ Media probably gets the profits for those. I'm not very good at economics. I think I'll stick with science.

--

Ash, Misty, and Brock trudged into the classroom for another day of lessons. Ash's face was still slightly bruised from the collision with the Pokéball.

"Man, you still look roughed up," said Richie, who was already in the classroom. "What did your mom say about the accident?"

"She was pretty upset," said Ash, "but she still made me come to school today. She says that this one incident doesn't mean that school is more dangerous than a Pokémon journey."

"I thought it was hilarious how Professor Oak just kept talking about centripetal force even after that happened," Brock mused.

"Well, you know what they say, the show must go on," said Misty.

Gary came marching into the classroom. "So, Ash, did you get in the way of any more flying Pokéballs?" he laughed.

"Why you…you wouldn't be laughing if you were the one that got hit," cried Ash.

"Well, I would have had the brains to get out of the way," said Gary.

Ash was ready to fight Gary, but at that moment, the bell rang and Professor Oak walked into the room. "Saved by the bell," whispered Brock.

"Good morning, class, I hope you didn't have any problems doing the homework from the last class," he said cheerfully. A.J. raised his hand.

"This is actually a question regarding the material from the last class, but you talked about the center of gravity between Ash and Misty, but I'm not entirely sure how you would calculate it quantitatively."

"That's a good question, A.J. I guess we can make it our first topic of discussion for today."

At that moment, Team Rocket came bursting through the door. Like the second day of class, they were beat up and out of breath, presumably from another failed attempt to enter the Ketchum residence.

"It's good for you to join us, and just in time for the first topic of our discussion. A.J. had asked us how about the center of gravity, and how to calculate it. The center of gravity is a point where the weight is affected by gravity. It is the same thing as the center of mass as long as the gravitational acceleration is the same the entire way through, because the acceleration would cancel out. Anyways, if you take something like a bottle of Full Restore and toss it, it would be pretty uneven because the Full Restore inside the bottle would slosh around. Even so, there would be a point that would trace out a normal, symmetrical parabola.

"Calculating the center of mass for an item where the mass is evenly distributed is pretty easy. It would be the center of the item. If you take a system where two objects with uniform masses are strung together, it is still fairly easy to figure out its center of mass. You would start by picking a point _x_, which can be anywhere from the middle of the system to an edge of the system. You would multiply the masses of each individual item by the distance the center of the item is from the point _x_, so it would be like m1x1 + m2x2. You take this sum, and divide it by the sum of the masses. The value you get is the center of mass relative to the location of the point _x_. So the answer you get will always change depending on where you set as an _x_. Be very careful about this when you do problems related to center of mass.

"Center of mass is important, but you usually wouldn't have to calculate centers of mass, because the calculation can become very complicated very quickly. One quantity you should be concerned about is Energy. Energy has a unit of Joules, and it can come in many different forms. We'll talk about most of them through the duration of this course, but for now we'll be concerned with two specific types of energy: potential and kinetic.

"The potential energy is the energy of an item based on its position. It is called potential energy because it refers to the object's potential to change its energy, or to do work. The potential energy is equivalent to the weight of the object times the object's distance from a certain height, or mgh. Let's say I take this Pokéball that weighs 1 kg and hold it here, 2 meters from the ground, what would its potential energy be?"

"20 Joules," said Todd. "And if you raise it even higher, to 3 meters, then its potential energy would be 30 Joules."

"That is correct," said Professor Oak. "The other major form of energy is kinetic energy, and this is the energy of an object in motion. The kinetic energy is equal to ½ mv 2, where m is mass and v is velocity. Let's take the Pokéball that hit Ash yesterday. Suppose it was going 3 m/s. What would be its kinetic energy?"

"The kinetic energy of the Pokéball that I would have been able to dodge so gracefully is equivalent to 4.5 Joules," Gary said proudly as his cheerleaders cheered behind him.

"Stupid Gary and his stupid cheerleaders," muttered Ash.

"Very good, Gary," said Professor Oak. "Potential and kinetic energies are very important, but the most important thing about them is that in all situations, the energy is conserved. It doesn't change. So if I throw this Pokéball up in the air and catch it, the total energy would be the same no matter where the Pokéball is. However, as you might suspect, the kinetic and the potential energies would be changing constantly. The maximum kinetic energy would be the instance after I throw it up, when the velocity is at a maximum. The maximum potential energy would be at the top of the flight, where the velocity in the vertical direction is zero.

"Conservation of energy is another way to solve for final velocity in free fall problems. Since the kinetic energy is zero at the top, and the potential energy is zero at the bottom, you can set the initial potential energy equal to the final kinetic energy: mgh ½ mv2. By doing a little rearrangement, you can figure out that the final velocity is equal to √(2gh). Notice how mass drops out in this equation. This makes sense, because we didn't have to deal with mass in any of our kinematics equations."

"I have a question," said Lara. "As we said last time, air resistance is a force, and this would make the final velocity slower than the initial velocity if you throw the Pokéball in the air and catch it. Would the energy still be conserved in this case?"

"That's a very smart question," Professor Oak said. "The energy is actually still conserved in this case, except that potential and kinetic energies are not the only energies present. The friction caused by air resistance would lead to the conversion of the energies into an entirely different form of energy, such as heat or sound."

"Oh, I see, so this complicates the equation," said Lara.

"Yes, it's good to keep this in mind, but you wouldn't have to worry about it in our tests. I'll be sure to tell you to ignore air resistance.

"There is a way to tie energy in with forces, which we covered last time, and that is with the work-energy theorem. I have already told you that work is the change in kinetic energy. Well, it is possible to define work as the force times distance times the cosine of the angle between force and distance: F d cos(θ). It is important to keep the cosine of the angle in mind, because if you want to move a Pokémon, but you're pushing down on it, then you're not doing any work at all, because angle of the desired distance and the force is 90°, and the cosine of 90° is zero. Work in this sense is different than what you would normally think of as work. For example, if you managed to pick up a Snorlax and carry it 5 meters, then you still didn't do any work, because the force you're putting in is against the weight of the Snorlax, which is along the _y_ axis, but you're moving the Snorlax along the _x_ axis."

"That is dastardly," said Brock.

"But you did work when you picked up the Snorlax in the first place, right?" asked Joe.

"That is correct, Joe. It is also possible to have negative work. If somebody pushes a Snorlax at you on a frictionless surface, then you're going to have to put force on it to get it to stop. Your force is on the opposite to the movement of the Snorlax, so the angle is 180°."

"And the cosine of 180° is -1, so you get negative work," said Tracey.

"That is correct," said Professor Oak. "Another quantity related to energy is Power, which is the change in energy over change in time, just like how velocity is change in displacement over change in time, and acceleration is change in velocity over change in time."

"That's the same as saying work over change in time, since work is the change in energy," said Misty.

"That's right, Misty. One last concept about work we'll cover before switching gears is a practical usage of work: pulleys. With pulleys, you can reduce the amount of force you need to lift a heavy object, like a Snorlax."

"We tried that once already, and it didn't work," griped Jessie. "We tried lifting a stupid Snorlax with ropes, and then the ropes snapped!"

"Ah, this must mean that your rope must have a low Young's modulus, and so reached the strain limit with limited stress. That will be a discussion for another day. Theoretically, if you have a strong enough rope, then you should be able to lift a Snorlax with relative ease.

"Pulleys work like this. Anything you want to lift up would have a force to it, the force of gravity. You can lift the item up with a pulley because the tension of the rope would offset the weight, and the item would accelerate upwards. The thing about tensions is that if you have a rope connected to a pulley that splits the rope in half, the two halves would have the same tension. If you hold the item in equilibrium, then the tensions from the rope would be equivalent to the weight. However, if there are two halves connected to the rope, it would mean the tension for each section would be half of the weight of the item, and the force you need to exert is equal to the tension of only one half of the string, which is half of the weight. If you have a pulley system that splits a rope into six pieces connected, then you'd only need to exert 1/6 of the energy.

"One thing you have to keep in mind about pulleys is that it is never 100 efficient. You always have to put in a little more work than you have to because some of the work is lost as heat or sound. The efficiency of the pulley is the ratio between the work you get out over the work you put in. Work is force times distance, so the work you get out is equal to the weight of the load times the distance you move the load, and the work you put in is your effort times the distance you move the rope. The effort is the actual force you put into the pulley, which is usually more than the expected force.

"In most pulley problems, you'd be given the effort and be asked to find the efficiency, or you would be given the efficiency and asked to find the effort. Either way, you need to use the efficiency equation, and interchange the efficiency and the effort.

"Our final concept is another concept that uses forces and the kinematic quantities: momentum. Momentum is a fairly straightforward topic, but it is very important. An object's momentum is its mass times its velocity, and it has the unit of kgm/s. For example, the Pokéball in yesterday's accident would have a momentum of 3 m2s, since it has a mass of one and a velocity of 3 m/s.

"The primary usage of momentum is in collision problems. There are two types of collisions: elastic collisions and inelastic collisions. While no collision can be truly elastic, it is still helpful to know the theoretical definitions of these concepts. An elastic collision is a collision where the kinetic energy of the items in the collision is conserved. The kinetic energy is not conserved in an inelastic collision, and in a completely inelastic collision, the two objects stick together and stay stuck together.

"Whether or not kinetic energy is conserved, you can figure out the final velocities of objects in a collision, because momentum is conserved as long as there are no external forces. In most problems we do, we'll assume momentum is conserved. This means, the sum of the momentum before the collision would be the same as the sum of the momentum after the collision. In most collisions, the equation would be m1v1,i + m2v2,i m1v1,f + m2v2,f. In a completely inelastic collision, the equation is m1v1,i + m2v2,i (m1 + m2)vf."

"As an example, let's take yesterday's collision between Ash and the Pokéball. Providing that the Pokéball is 1 kilogram and moving at 3 m/s, and Ash is 40 kg and starting from rest, what would be the final velocities in an inelastic collision and an elastic collision?"

"Well, in an inelastic collision, the Pokéball would stick to Ash and give him a miniature whiplash," said Tracey. "The initial momentum for the Pokéball was 3 kgm/s, with a mass of 1 kg and velocity of 3 m/s. Ash's initial momentum would be 0 kgm/s. Once they stick together, they would have a mass of 41 kg, but their momentum would still be 3 kgm/s, so their final velocity must be 3/41 m/s."

"Very good, Tracey. What would the final velocity be in a completely elastic collision?"

"That's easy!" cried Gary. "The Pokéball's initial momentum is 3 kgm/s, and its initial kinetic energy would be 4.5 J. The final momentum of the Pokéball and Ash would add up to be 3 kgm/s, and the final kinetic energy of them would add up to be 4.5 J. The Pokéball is 1 kg while Ash is 40 kg. Putting it all together, this would mean that the final velocities are…" He paused and frowned. He began wracking his head for an answer and finally said, "You know, I know what to be looking for, but I can't seem to find the equation to figure it out."

"Ah, yes, that's okay," said Professor Oak. "I didn't expect any of you to figure it out. The fact that the kinetic energy is also conserved means that only one pair of velocities would work."

"Well, the derivation was tricky, but it didn't require any complex calculus, so I think I figured it out," said Misty. "We have two equations at our disposal, the conservation of momentum and the conservation kinetic energy. I can solve one and insert it into the other equation. In this case, I solved for Ash's final velocity, equal to (m1/m2)(vi-v1f), and put it into the kinetic energy equation. I ended up with a quadratic equation where the Pokéball's final velocity is the _x_: m1 (m1 + m2) v12 + -2m12v1 + m1 (m1 + m2) vi2. Since it was a quadratic equation, there would be two answers, (m1 + m2) / (m1 + m2) vi v1 and (m1 - m2) / (m1 + m2) vi v1. The former equation is useless because it's equal to 1, but the other equation would be sufficient to tell us the final velocity of the Pokéball. After that, I just put this value for the Pokéball's final velocity into the equation for Ash's final velocity to get an equation with only masses: v2 2m1 / (m1 + m2) vi."

"Based on these equations, the Pokéball's final velocity is (-39/41) 3, or -2.85 m/s, or 2.85 m/s in the opposite direction," said Tracey. "Ash's final velocity is (2/41) 3, or .15 m/s."

"Uh, I didn't get any of that," said James.

"It's okay," said Professor Oak. "I didn't expect you to have to do any of the messy algebra related to this derivation, but it's helpful to understand how to work with the equation like what Tracey did.

"Finally, momentum can be related to forces through the concept of impulse. As you know, a change in the net force can lead to an acceleration, which would lead to a change in velocity…"

"And a change in the momentum," said Todd.

"That is correct. This change in momentum is the impulse, and it is equivalent to a net force applied for a certain amount of time. For example, if you apply a force of 10N for 10 seconds to a Pokéball, it would have an impulse of 100 kg m/s."

"And the change in the momentum would be 100 kgm/s," said Duplica. "If the Pokéball is 1 kg, then its velocity would change by 100 m/s."

"That's one fast Pokéball," Meowth remarked.

"The relationship between force and time is extremely important when it comes to car accidents or falls from a large height. If you're falling from a building and keep your leg straight, then you'll most likely break all your bones in the body. However, if you bend your legs a little bit, then the damage may not be so bad. Can anybody explain why?"

Richie answered. "Well, assuming the momentum right before impact is the same in both cases, then the force you feel and the time is inversely related. If you keep your knee straight, you'll have a short moment of impact, and the force you'll experience is high. However, if you bend your knees to the impact, then the moment of impact would increase, and the force you'll experience would decrease."

"Oh, so that's why Mondo's always got that stupid trampoline of his whenever he comes to rescue us after we blast off," said Meowth. "The trampoline bends, and that decreases the impact time, thus absorbing the force."

"Come to think of it," said James, "he must be really good at physics, considering he always knows where we land."

"You see," Professor Oak said, "it's always good to have a solid understanding of physics, because it just may save your life. One last topic that we will talk about is angular momentum. You should already know that there is a rotational equilibrium defined by torque that is similar to our usual translational equilibrium. Well, there is also a momentum based on how an object spins. The angular momentum is very similar to translational momentum. There is just one difference. While translational momentum is mass times velocity, angular momentum is mass times velocity times the radius of the object's spin.

Professor Oak took a spinner from behind the desk along with two large weights. "There's a common demonstration to help people understand angular momentum. Who would like to volunteer?" Most of the class raised their hands. In the end, Professor Oak chose Ash. Professor Oak told Ash to step on the spinner and hold the weights as far apart as he can. Afterwards, he began spinning the spinner. Once Ash began spinning with a certain speed, Professor Oak told him to bring the weights as close to him as he can. Ash did so, and began spinning at a much faster speed.

"You see? The angular momentum measures the momentum of the spin. Ash was originally spinning quite slowly, since the radius of his spin was large because he held the weights were so far out. Once he took the weights close to him, the radius shrank. Since momentum is conserved because there weren't any external forces, his angular velocity would increase dramatically."

"Thank you, Ash," he said once Ash stopped spinning. Ash may have stopped spinning, but his semicircular canals were still going. He took a couple of steps before falling flat on the ground. Most of the class broke out into laughter. "Nice going, you klutz!" laughed Gary.

Professor Oak told the class to quiet down before he went on. "All right, that's it for today's class. We'll be switching gears a little bit for tomorrow as we start our discussion as the other important physical science: chemistry. Be sure to do the problems related to work, energy, and momentum, and be ready to learn about atoms and nuclei."

The bell rang, and the students began scrambling out of the room. Ash stayed behind as Misty and Brock went to speak with Professor Oak, and silently cursed Gary. "Someday I'll get you in this class, just you wait," he said.

--

AUTHOR'S NOTES: I think by now I've started to make it a goal to end every odd chapter with "This is going to be a long year," and every even chapter with "Someday I'll get you, Gary!" or something close to that. I have no idea how long I can keep it up. There's not much else to say about this chapter except I hate Tension, and I made my life miserable by coming up with an example that required the derivation of the equation for elastic collisions. I had to look it up online, and I still don't completely understand it. That's what happens if you ignore the math you learned. I had to add the thing about angular momentum in afterwards, but it's a was good demonstration to make Ash seem more pathetic. XD


	5. Chapter 5: Atomic Structure

Brrring

DISCLAIMER: I don't own anything in this story except for the ideas, but I don't even think that belongs to me. (Well, if you think about it, ultimately everything belongs to God, including our own bodies. Which is why the Bible implores us to "treat our bodies in a way that is honorable and holy, not in passionate lust like the heathen who do not know God" 1 Thessalonians 4:4-5. I think I'm writing this more for myself than anything else.) Nintendo owns the rights to Pokémon, and Shogakukuan owns the Pokémon anime…I guess only in Japan. I think Pokémon USA now owns the rights to the anime in America, and that's a subsidiary of Nintendo anyways.

--

School was ready to start, and Ash came into the classroom looking more determined than ever.

"So, did you do your reading about atoms and nuclear phenomena?" asked Brock.

"No, I was thinking of ways to get Gary back," Ash replied.

"And did you come up with anything?" Misty wondered.

"Not a thing," Ash said candidly.

"I'm not surprised."

Richie spoke up. "Well, if you want to get the best of Gary in the classroom, perhaps it would be best to get a good handle of the topics."

"I try, I really do," Ash complained. "But all this stuff is so boring in comparison to Pokémon training."

"Well, perhaps you should just drop out, but don't blame me if we become much better trainers than you because we continued our studies," said an obnoxious voice in the back. It was Gary.

"You only wish!" cried Ash. "I'm going to be a zillion times the trainer than you'll ever be!"

At that moment, the bell rang. Ash sat down in rage, mumbling and grumbling about Gary. Professor Oak came walking in, but at that moment, Team Rocket came in. The four of them collided, and they ended up sprawled on the floor.

"I'm so sorry about that," said James, "but we were just trying to get into the house and capture…"

"Shut up, pea-brain!" cried Meowth. "You'll give away our plans!"

"It's all right," said Professor Oak. "In fact, that collision is a perfect illustration of the sort of physics that we've been going over the past few days. Would anybody care to try an explain it?"

Misty waited to see if anybody else would answer. When nobody did, she spoke up. "It's very simple. You were walking at a certain velocity, and Team Rocket came barreling into you at a much higher velocity. After the collision, which we will say was inelastic, you ended up going at a speed that was faster than your initial speed, because Team Rocket transferred their mass and their speed to you. However, your feet didn't get the message, and it was still going at your original pace. We all require static friction to walk, and because of the static friction, your feet strayed behind the rest of your body. Soon, the center of gravity went far beyond the point of support, and you ended up crashing to the floor."

"That's very good," said Professor Oak. "Here's another question, was my momentum conserved in that collision?"

"Sure it was," said Gary. "After all, momentum is always conserved in the absence of an external force."

"That may be true," said Duplica, "but you must recall that Professor Oak was talking about his momentum, which means that Team Rocket's momentum isn't figured in this equation. Since the professor's velocity changed, his momentum changed even if the system's momentum is conserved."

Gary was aghast. "That's right! And there was an impulse. The moment of impact was quite short, which means the force that gramps felt was high! All right you," he said to Team Rocket, "if anything happens to my grandpa, you're going to get it!"

"Just you try," said Meowth, bearing his claws.

"All right, there will be no fighting in the classroom," Professor Oak admonished, "but I thank you for your concern, Gary. Anyways, today we'll be making a little bit of a transition and begin our discussion on chemistry. Where physics is the study of the physical world, chemistry is the study of matter. Everything is considered matter: you, me, this Pokéball, and all of the Pokémon in the world, even the legendaries. Who here knows what matter is made of?"

Lara raised her hand. "All matter is made up of atoms."

"That is correct," said Professor Oak. "Atoms are a fundamental part of all things. This Pokéball and the air we breathe are all made of atoms. Since it's so important, we need to have a basic understanding of the structures and properties of.

"You can think of atoms to be like the Solar System, even though it is far more complicated than that. There is a nucleus in the center made up of protons and neutrons. This is like the Sun. There are many electrons orbiting this Sun, and these are like the planets. The protons, neutrons, and electrons are considered subatomic particles, and they are the basis of an atom's physical and chemical properties. Some of you know may know that Oxygen and Nitrogen gas are the primary components of air, while Carbon is the basic component of coal, diamond, and pencil lead. These elements may seem have extremely different properties, but they by just a few protons in their nuclei.

"The identity of an element is based on its atomic number, which in turn is dependent on the number of protons in their nucleus. Hydrogen has only one proton, and it has an atomic number of one. Carbon has six protons, and it has an atomic number of six. Oxygen has eight protons, and it has an atomic number of eight. Another important property of an element is its atomic weight, which basically tells you how much an atom weighs. This is dependent on the number of protons and neutrons, since they are about equal in mass. The basic unit of atomic mass is the atomic mass unit, or amu."

A.J. raised his hand. "I have a question. I'm looking at this table of the elements, and most of the elements have atomic weights that are decimals. For example, Cl has an atomic mass of 35.5. Does this mean that each atom of Cl has 35 and a half protons or neutrons?"

"That's a very good question," Professor Oak answered. "Actually, the protons and neutrons atoms have are all whole numbers, and while the number of protons may stay the same in an element, the number of neutrons they have may change."

"Oh, I see. So there are different atoms of the same element with different weights."

"Yes, these are known as isotopes. There are only a limited number of isotopes for a given element that are stable and can exist naturally. For example, Chlorine has an isotope that's 35 amu, and another that's 37 amu. Since there are so many atoms in the world, you never know how much of which type you'll get in a sample, so the atomic masses are given as a weighted average of each of the common isotopes. 75 of the world's Chlorine is 35 amu, and 25 of them are 37 amu. If you do the mass, you'll find out that the weighted mass is 35.5 amu."

"Another important property of these subatomic particles is that they carry a charge. Ash, you of all people should know about charges, since Pikachu is your primary Pokémon. The charges held by the subatomic particles are exactly how Pikachu can do its shocking electric attacks."

"Those stupid charges," muttered Jessie.

"We'll be discussing electrostatics later, but it's important to understand how charges work in these subatomic particles. Protons have a positive charge. Electrons have a negative charge, and neutrons have no charge. They are neutral. Most atoms don't like to have a charge, so the number of electrons usually matches the number of protons, so Hydrogen would have one electron; Oxygen would have eight and so on."

Todd had a question. "You know how there's the convention that like charges repel while opposite charges attract? So how come the protons are able to stick together in the nucleus without a problem?"

"That's a very good question, Todd. The truth is the repulsion of like charges is based on electrostatic forces, which we'll learn about later. However, there is a force called the strong nuclear force that can overcome this electrostatic repulsion, but only in extremely small distances. Once the nucleus gets to be very large, the nuclear force may not be sufficient to keep the nucleus together. The effect of this is known as radioactive decay. In radioactive decay, atomic nuclei would begin losing its protons and neutrons, and become smaller nuclei. There are four common decay particles.

"The alpha particle is when a large nucleus emits a Helium nucleus, which contains two protons and two neutrons. A nucleus that emits an alpha particle would lose two atomic numbers and four amu from their atomic weights. The second type is the beta particle. Beta particles are essentially electrons or their opposites, the positron. When an electron is emitted, this is known as beta- decay, and it happens when a neutron decays to become a proton. When this happens, an electron is ejected from the atom. The opposite is beta+ decay, and this is when a proton decays to become a neutron and a positron. Who can tell me what happens to the atomic mass and the atomic number in beta decays?"

"The atomic mass wouldn't change," said Duplica, "but in beta- decay, the atomic number would increase by one. In beta+ decay, the beta decay would go down by one."

"Now I find that to be quite confusing," said Samurai. "One would expect that in beta- decay, the atomic number would go down because of the negative sign, and vice versa."

"I can see why you're confused," said Misty, "but just think about it. The signs refer to whether or not an electron or a positron was emitted. However, the important thing to keep in mind with the atomic number is what happens in the nucleus. In beta- decay, a neutron becomes a proton. In beta+ decay, a proton becomes a neutron."

"I see," Samurai replied. "I suppose it would be helpful to think of it as opposites. The atomic number would become more positive in beta- decay."

"Well, I guess you'll have a really hard time when we come to talk about ions. Anyways, a third type of decay is gamma decay, which is when a large nucleus emits a gamma particle. Atoms emit electromagnetic radiation all the time, as we will eventually learn, but gamma rays are extremely high energy and can be life-threatening. The only thing you really have to know is that nothing happens to its atomic mass or its atomic number. Finally, there is something known as electron capture. This is rare, but it happens when a large nucleus captures an electron. This is the opposite of beta- decay, and rather than a neutron breaking up to form a proton and an electron, the electron combines with a proton to form a neutron.

"All elements decay randomly, which means in a sample of particles, you never know which nuclei would decay. However, the rate at which a sample of elements decays is a constant, and this is known as the half-life."

"Oh yeah, I love playing _Half-Life_! Gordon Freeman is such a stud," said Lara.

"And don't forget about _The Orange Box_. _Team Fortress_ and _Portal_ are great too," said A.J.

"Ahem," said Professor Oak, and the two students were quiet. "The half-life is the amount of time it takes for half of the nucleus decay. So let's say you have 16 nuclei of an element, and the half life of the particle is 5 days, how much time would it take to leave only one nucleus?"

Richie did the math. "Well, after 5 days, there would be 8 nuclei remaining. After 10 days, only four would remain. So it would take 20 days for three of the four remaining nuclei to decay."

"It's weird how the rate of decay would just slow down like that if you think about it this way," said Tracey.

"It is weird, but that is the mysteries of science. You can mathematically how much of a nucleus is left with the equation n noe-λt, where no is the original number of nuclei, e is the base for the exponential function, λ is the decay constant, and t is the half-life."

"Geez, it's so weird seeing λ as the decay constant, since I've come to associate that symbol with the half-life because of the game," A.J. complained.

"Well, get used to it," said Professor Oak. "The decay constant λ is actually very closely related to the half-life. It is equivalent to the natural log of 2 / the half-life. The natural log of 2 is .693, so λ .693/t.

"Another important aspect of nuclei is the nuclear binding energy. I'm sure you've all heard of the phrase E mc2. Well, that equation is related to this concept. Whenever the nucleon – that is the protons and neutrons – come together to form the nucleus, the mass of the nucleus is slightly different than the individual masses of the nucleons. This difference in mass is known as the mass defect.

"Oh, I know all about _Mass Effect_," said James. "I love playing it on the Xbox 360."

"Not _Mass Effect_, you idiot, mass defect!" cried an exasperated Meowth.

"Wait a minute," said Brock, "we're all Nintendo characters, so why are you playing an Xbox 360?"

"We're villains! What else do you think we play?" said Jessie.

"We love the PlayStation 3 too," said James.

"Now if we'll get back to the topic at hand," said an impatient Professor Oak, "the mass defect is the _m_ in that famous equation. The _E_, of course, refers to the nuclear binding energy. The higher the binding energy is, the more stable the nucleus.

"These are the basic nuclear phenomena. The real fun about atoms comes when we talk about electrons. Electrons are a completely different beast. Up to now, we have been learning about classical Newtonian mechanics, but the mechanics that govern these electrons are known as quantum mechanics, and they are quite different from Newtonian mechanics.

"The first model of an atom was formulated by Niels Bohr, who came up with a model of the hydrogen atom, which has one proton and one electron. Using the angular momentum concepts we talked about last time and theories made by Max Planck saying that light can only have given energies, Bohr postulated that an electron would have set orbits around the proton with energies equal to –RH/n2, where RH is a constant known as the Rydberg constant, and n is the orbital the electron is at.

"The model also states that if a particle of light – known as the photon – of certain energy strikes the electron, then the electron would absorb the photon and be excited to a higher orbital. Once the electron goes back to a lower state, it loses energy and re-emits the electron. This leads to an absorption spectrum or an emission spectrum respectively, and scientists can use these to fingerprint elements. We'll talk about these energies and how they relate to Bohr's model later, after we explore electromagnetism.

"Bohr's model is good for Hydrogen and other atoms with only one electron, but it falls apart when we get multiple electrons. Electrons move too rapidly to be detectable, and we can never figure out where exactly an electron is. This is the basis of Heisenberg's uncertainty principle. However, electrons still have set spaces it can spin around known as energy levels. We can identify an electron, knowing on which orbital it spins through a series of quantum numbers.

"The principle quantum number, or _n_, determines the size of the energy level. This is similar to Bohr's idea that the electron can only occupy levels of certain energies. The higher the _n_, the higher energy is contained at the level. You can figure out the maximum number of electrons in an energy level of size _n_ with the formula 2_n_2. The next quantum number is the angular momentum, or the azimuthal quantum number _l_. This specifies the subshell of the energy level is at, and also the shape. The azimuthal quantum number for a given _n_ is 0 to _n_ – 1, so an energy level with the number 4 would have subshells 0, 1, 2, and 3. These subshells all have names. 0 is s, 1 is p, 2 is d, and 3 is f. Any nucleus with naturally occurring subshells beyond f is too unstable to be detected. The maximum number of electrons in a subshell can be calculated with the equation 4_l_ + 2.

"The magnetic quantum number _m__l_ is the specific orbital in a subshell the electron is at. The given orbitals for each subshell can be numbered –_l_ to _l_, so a p subshell, which as an azimuthal quantum number of 1, can have orbitals -1, 0, and 1. The maximum number of electrons in a given orbital is two, and each of these electrons has a different spin quantum number _m__s_. The _m__s_ value of an electron can be +½ or -½."

"So is it physically possible for two electrons to have the same four quantum numbers?" asked Todd.

"Well, if the electrons are in different atoms, then yes, but if they're in the same atom, then no. Each electron in an atom has different quantum numbers. This is known as the Pauli Exclusion Principle. If you think about it, the quantum numbers are like an address. The principal quantum number is like the city the electron is in, like Cerulean City or Pewter City. The azimuthal quantum number is like the street the electron lives on. The magnetic quantum number determines the building on the street, and the spin quantum number determines the room where the electron lives."

"That city is quite conformed," Samurai remarked, "similar to the cities in the novel _A Wrinkle in Time_."

"If you think about it, it is," replied Professor Oak. "Whenever you add a proton, then it's pretty natural for an electron to come on, and electrons like to occupy positions with lower energies. There is a rule known as the Aufbau Principle that determines the order of the subshell that electrons occupy. To find the subshell that fills first, you do the simple math (_n_ + _l_), and the lower values fill up first. If they both have the same (_n_ + _l_) value, then the subshell with the lower _n_ fills first. Another thing about electron filling is known as Hund's Rule. Let's say you have three p orbitals that can hold six electrons, but you only have three. How would you fill up the electrons?"

"Well, obviously you'd fill up one orbital and put one in another," said Ash.

"Actually, electrons prefer lower energies. It takes up more energy to put an electron into an orbital that already has an electron in it than to put electrons in empty orbitals, so if you have three electrons, they'd fill up in the three orbitals and have parallel spins," said Richie.

"Yeah, Ash, you stupid," said Gary.

"Argh, stupid Gary!" cried Ash.

"Maybe you should just keep your mouth shut," said Misty.

"Richie is correct. Hund's Rule states that electrons would partially fill the empty orbitals in a subshell before going on to the other orbitals. This is why Nitrogen is more stable than Carbon or Oxygen, because it has half-filled orbitals. However, the most stable atoms are the noble gases, which have their orbitals completely filled.

"Finally, there's the concept of valence electrons. Valence electrons are electrons that are in the outermost shells, and are thus available for bonding. The number of bonds an atom can make is related to its valence electrons. This will become a very important concept in our discussion for tomorrow, when we talk about bonding and the periodic table. For now, have a good rest of the day. Be sure to do the homework for this chapter, and the reading for the next chapter."

The bell rang, and the students began leaving the classroom. Ash stayed behind. The humiliation was just too much for him. Richie looked at his friend sadly, and said, "This is going to be a long year."

--

AUTHOR'S NOTES: This chapter has probably the best thing that I'll ever write in this story: the little snippet about Mass Effect. It's probably not even that good, but you can definitely see Team Rocket break out the X-Box Live to play GTA IV or Call of Duty 4. The mentions of Half-Life only a few lines earlier probably blunts the effect of the video game reference, but if you're a gamer you can't learn about half-life without thinking about the game. The citation of The Orange Box may be a bit excessive, though. As for the discussion of momentum early in the chapter, it was something I came up with as I was thinking of ways to fill up space before class starts. It's probably the only time that'll happen. It's interesting because most physics classes leave decay until the end, but I switched things around just so it can be integrated with the chemistry.


	6. Chapter 6: Periodic Table and Bonding

Brrring

DISCLAIMER: I did a little research on the ownership of Pokémon, and Nintendo does own everything. Pokémon USA is responsible for marketing and licensing. Shogakukan (which I've been spelling wrong for a long time) is responsible for the distribution of the anime and manga in both Japan and America. Anyways, they're both giant corporations, so don't mess with them.

--

Ash Ketchum walked into the classroom for yet another day of school, and rather than sit at his usual seat near the front of the room but also right in front of Gary, he quickly moved to the other side of the room to be with Richie. Gary watched him with an amused smirk. "Moving to the other side of the room doesn't hide the fact that I'm so much better than you in all of this," he cackled.

"Just ignore him," Richie consoled. "The only thing he wants is for you to react to him. Don't give him that satisfaction."

"I don't need Ash to do anything for me to feel satisfied, since I already know I'm so much better than him as a student and as a Pokémon trainer!"

"You know, there's a line between innocent banter and malicious insults, and you've crossed that line a long time ago," said Misty.

"You're ruining Ash's desire to learn and disrupting the class with your comments," said Brock. "That should be enough to get anyone kicked out of class."

"Aw, Gramps wouldn't ever expel me," Gary scoffed confidently. "And besides, I'm just making Ash tougher! I'm doing him a big favor!"

Richie turned to Ash and said, "Is this what you had to deal with every day when you were growing up?" Ash nodded solemnly.

The bell rang, and Professor Oak walked into the room. Ash knew that Gary wasn't his only worry. Team Rocket was still after his Pikachu. He wondered when his luck will run out, and his most loyal friend would end up being captive to the evil corporation. Team Rocket scrambled in at the last second mumbling and grumbling, so he knew he had another day.

"All right, today we will continue our discussion of basic chemistry," Professor Oak said when Team Rocket sat down where Ash, Misty, and Brock used to sit. "Our first topic of the day will be the periodic table. You've probably seen the periodic table every day in this class. Well, the periodic table resulted from hundreds of years of research into the elements and how to group them in a meaningful way. You should already know that an element's chemical and physical properties are based on the atomic number. Well, the periodic table will allow you to predict what the properties may be, and what other elements have similar properties.

"Just take a look at the table. The horizontal rows are known as the periods, and the vertical columns are known as groups. The elements are actually ordered based on when the electrons fill up based on the Aufbau Principle, which makes sense if you consider the fact that the electrons increase at the same rate as protons.

"The important thing to know about groups is that they are organized based on the number of valence electrons, which we talked about last class. Since valence electrons are an important determinant of activity, each element in a group has similar properties. Each group has a name based on the number of valence electrons. The elements in these left-most two columns and five right-most columns are known as the representative elements, and the groups have names with A, like IA, IIA etc. The elements in the middle are transition elements, and they are the elements with _d_ electrons. These groups have names with B. The left-most column of transition elements is actually known as group IIIB. Can anybody say why?"

"Well, they're the third group from the left, so they'll have three valence electrons," said Joe.

"That's very good. Determining the valence electrons for transition elements is a bit complex, but you should keep in mind that group IIIB is on the left, and IIB is on the right. Down beneath everything else are the _f _elements, the lanthanides and the actinides. These elements are quite rare, and you'll only experience them in this class through decay questions.

"Since elements in the same groups have similar properties that are different from elements in other groups, it's important to keep in mind the necessary groups. For example, elements in group IA are known as the alkali metals. These are elements with one valence electron. Since most elements would rather have a complete orbital, these elements are extremely eager to donate their electrons so they can have a complete orbital in the _n_ energy level beneath them. They are very reactive, and would burn once they come in contact with water.

"Elements in group IIA are the alkaline earth metals. They're reactive, but less so than the alkali metals. Group IVA is known as the Carbon group. They have four valence electrons, which is half of an energy level, so these elements are extremely versatile, especially Carbon. The second half of this course will deal with Organic Chemistry, which is the chemistry of carbon compounds. Group VA is the Nitrogen group. Elements in this group are slightly more stable than the others, because they have half-filled _p_ orbitals. Group VIA is the Oxygen group, and by the time you get here, these elements want electrons to have a full energy levels, so they'll be willing to accept electrons. We'll talk about these later. Group VIIA consists of the Halogens, and these are only one electron away from having a complete energy level, so they're extremely reactive as well. Finally, Group VIIIA is the Noble Gases. They automatically come with full energy levels, so they have very little desire to react with other elements. They are very stable.

"This custom of having eight valence electrons as a standard is known as the Octet Rule. The elements we deal with the most in chemistry all want eight valence electrons, although elements in the third period and lower can have more than eight valence electrons, because they have empty _d_ or _f_ orbitals that can be used for bonds.

"The elements on the left are known as the metals. They are the elements that typically want to give up electrons. They are characterized by good conductivity, a shiny exterior, and malleability. Hydrogen is the only element on the left that's not a metal. Elements on the right are known as non-metals. They have opposite properties from metals. They are poor conductors, no luster, and are brittle rather than malleable. In the middle of this periodic table, there is a line showing the separation between metals and non-metals. The elements bordering this line are known as metalloids. They have properties that are akin to metals and non-metals."

A.J. had a question. "So, I understand that steel is a type of metal. How exactly does it fit in with this metal and non-metal business?"

"Ah, steel is a special case," said Professor Oak. "Steel actually isn't an element, and it isn't completely a metal either. Steel is actually an alloy, which is a solid mixture of a metal with another element, sometimes another metal or a non-metal. Steel is an alloy of Iron and usually Carbon. Both of these are essential elements, so it would appear naturally in Pokémon. The Carbon gives steel a certain hardness that would explain the Pokémon's high defense. The Iron crystal lattice structure is able to block out all poison, which is really just different sorts of molecules anyways.

"There are several trends in the periodic table that are generally determined by the element's location, which is determined by everything from the size of nucleus to the number of electrons, and the highest principal energy level in the ground state. The most basic of the atomic trends, one that can be used to explain all of the other atomic trends, is the effective nuclear charge. This is the measure of the electrostatic interaction between protons and electrons. The higher the interaction, the higher the charge, and the harder it would be to pull off an electron. Generally, the more electrons are in the valence shell, the more protons in the nucleus, and the closer the valence shell is, the higher the effective nuclear charge would be."

"So the rightmost elements would have the highest effective nuclear charge, and the leftmost elements would have the lowest. And the higher the element is, the higher the charge would be," said Todd.

"That is correct. Once you consider everything in terms of this electron pull, then the other atomic trends should come naturally. For example, the ionization energy is the energy that must be put in to remove an electron. Naturally, the more protons an element has in its nucleus, the higher the electrostatic pull, and the more energy you must put in to remove this pull. There are different ionization energies, such as first and second ionization energies, depending on how many electrons you remove. It typically takes even more energy to pull a second electron away, because it would give an atom an even more positive charge, so second ionization energies are generally much higher.

"The electron affinity is the opposite of ionization energy. It measures the energy released once an element gains an electron. Even though the release of energy is given by a negative sign, the convention is that the more energy is released, the higher the electron affinity. Some elements have positive energies, so you must put in energy for the atom to accept the electron. This is equivalent to a very low electron affinity."

"It's like having to pay somebody to take a Pokémon item rather than other people paying to get a Pokémon item," said Lara.

"That's a very good analogy, Lara. A third periodic trend is electronegativity. If you think about the effective nuclear charge, the more valence electrons an element have, the more they want more electrons. So when two atoms come together in a bond, which we'll talk about later today, one atom may have a stronger pull on electrons than the other. The higher this pull is, the higher the electronegativity is. If you think about these three atomic trends, they follow the same trend as the effective nuclear charge. They decrease as you go from right to left, and from up to down.

"The final periodic trend is the opposite. It is the atomic radius. If you think about it carefully, the higher the electrostatic pull is on the electrons, the closer the valence shell electrons would be to the nucleus, and the smaller the radius would be. Therefore, as you go more to the right, the effective nuclear charge will increase, and the atomic radius will decrease. And as you go from up to down, the principal quantum number would increase. The principal quantum number measures the size of the energy levels, so the atomic radius would increase as well. So what would be the largest atom at the smallest atom?"

"The largest atom is Francium, and the smallest is Helium," said Duplica.

"That is correct. So what exactly is the point of understanding the properties of atoms? Well, it's so we can delve into the meat of chemistry, the concept of binding. As you know, atoms rarely ever exist as separate entities, but they combine with other atoms to form molecules. The reason they do this is because almost all of the elements have partially-filled orbitals, and the goal of almost all of them is to have full orbitals. Well, these chemical bonds are how atoms achieve full orbitals. The Octet Rule that I talked about early is the typical guidelines to see whether or not an atom is satisfied, although there are more exceptions to the rule than elements that full it to the dot. The Octet Rule basically states that elements are happy with eight valence electrons.

"There are two basic types of bonds. Which bond two atoms would use depends on difference in electronegativities. If two atoms differ in electronegativity by more than 1.7, they would bind in an ionic bond. This 1.7 electronegativity difference is not an absolute law, but it serves as a good rule of thumb. If they differ by less, then they would bind in a covalent bond. You don't really need to figure out electronegativity values to figure this out, though. Any time you have a representative metal and a non-metal, it's almost guaranteed that they'll bind in an ionic bond. In an ionic bond, the non-metal would strip off the electrons of the metal, creating two ions with complete octets, a positive cation and a negative anion. As you know, opposite charges attract, and these two ions would combine electrostatically. And these ionic bonds aren't one and done events either. A typical ionic interaction between two atoms would result in large crystals made up of more than 1 x 1023 atoms. An example of an ionic crystal is table salt, which is simply Sodium and Chlorine.

"The other type of bond is where most of the chemistry we'll look at occurs. It's the covalent bond, and it usually happens between two elements with similar electronegativities, typically between two non-metals or Hydrogen. The atoms are not electronegative enough to completely strip off the other atom's electrons, but they generally come together and share the electrons, so that some electrons would spend time on one atom and then on another atom. Two atoms can have as many as one, two, or three pairs of shared electrons based on how much they need to complete the octet. The number of these bonds is known as bond order, and bond order is tied together with bond length and the bond energy, which measures the energy needed to break a bond. The bond length and the bond energy are inversely related, so single bonds have the longest bond length but also smallest bond energy. Triple bonds are the shortest and the strongest.

"There are three main types of covalent bonds, and these are generally related to electronegativity differences. For two elements with different electronegativities, even though the more electronegative element isn't strong enough to strip off the other atom's electrons, they would still have a stronger pull on the elements, and so the electrons would spend more of its time with the more electronegative element, creating a dipole, with partial negative charge on the more electronegative end and a partial positive charge on the other end. This is known as a polar covalent bond, and would create polar molecules if the dipoles don't cancel out. Water is polar because two dipole moments exist between Oxygen and Hydrogen, and these dipole moments point in a certain direction. Nonpolar covalent bonds happen when atoms have the same electonegativities, or have very similar electronegativies, like Carbon and Hydrogen, so the hydrocarbons we'll learn about in organic chemistry are non-polar.

"The final type of covalent bond is the coordinate covalent bond. These are generally quite rare. Usually in covalent bonds, each of the atoms would donate an electron to the shared electron pair. However, with certain molecules, both of the electrons in an electron pair came from the same atom. A very common example is with Lewis acids and Lewis bases, which we'll talk about in a later course. A Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor.

"We can also take a look at covalent bonding on an orbital level. The bonds we look at involve _s_ and _p_ orbitals. Orbitals exist separately in atoms, and come together to create new molecular orbitals when you they combine to form molecules. As you know, the _p_ orbitals have magnetic quantum numbers of -1, 0, and 1. If two orbitals of similar signs combine together, then they form a more stable bonding MO. If two orbitals of different signs come together, they form a less stable antibonding MO. These molecular orbitals can come in head to head connections, or when two p orbitals are parallel to each other. Direct connections between orbitals result in σ (sigma) bonds, while parallel orbitals result in π (pi) bonds.

"All single bonds are σ bonds, while double and triple bonds have one or two π bonds. As you would probably expect, π bonds can't exist without σ bonds. Recently, experimental data shows that the atomic orbitals would change shapes whenever they come together in σ bonds to create a new MO with properties of the orbitals that make up the bond. This is called orbital hybridization. In a molecule with four single bonds, the _s_ orbital and three _p_ orbitals come together to form four _sp__3_ molecular orbitals. When there is a double bond between two of the atoms, then a _p_ orbital is disposed, and you are left with three _sp__2 _orbitals and a π bond. Similarly, in a triple bond, you have an _s_ and a _p_ orbital to create two _sp_ orbitals with two π bonds.

"Depending on the hybridization, you have different bond angles to maximize the distance between bonds. This is known as the Valence Shell Electron-Pair Repulsion theory, or VSEPR. With VSEPR, you can determine the bond angles and the shapes of these atoms. The most basic shape comes with a _sp__3 _molecule with four bonds. This is known as the tetrahedral molecule, and they have bond angles of 109.5°. However, sometimes an element like Oxygen or Nitrogen would have an extra electron pair that is not involved in bonding but still occupies an orbital. This would result in a trigonal pyramidal molecule if there is one electron pair and a bent molecule if there are two electron pairs. An example of a bent molecule is water. Electron pairs take up more room than bonds, so the bond angle in water is smaller than a normal tetrahedral bond angle: 104.5°.

"A molecule with a double bond has _sp__2 _orbitals. These molecules are flat, and the other atoms are arranged in a triangle, creating what is known as trigonal planar, since the molecules occupy one plane. The bond angles in these molecules are 120°. However, if the central atom has an electron pair, then there would only be two atoms bonded to it. This is also a bent molecule. Finally, a molecule with a triple bond can only have two things bonded to each other, and so this is considered a linear molecule with 180° bond angles. We will deal with some molecules that break the octet rule, so it is important to learn these structures as well. If a central atom in a molecule has five atoms bonded to it, then this creates a diamond shape with three faces. This is trigonal bipyramidal, since it's like two trigonal pyramidal molecules piled on top of each other. This has bond angles of 90°, 120°, and 180°. Finally, if a central atom has six atoms bound to it, then it creates a diamond with four faces. This is called octahedral, and they have bond angles of only 90° and 180°. Draw the molecules to see if you can see the bond angles.

"If you have no artistic skills and nevertheless want to draw molecules, then you would have to draw Lewis structures. In these Lewis structures, you can depict 3D molecules on 2D planes and get away with it. The first thing you should do is count the number of valence electrons there are, so in a molecule like water, there would be eight: six from Oxygen and one each from the two Hydrogens. Next, you should put the least electronegative atom that is not Hydrogen in the middle. In this case, it's Oxygen, but usually you'd have something like Nitrogen or Carbon. Next, you should draw a single bond from the central atom to each of the other atoms. After that, you should draw two dots for electron pairs around the remaining atoms until the octets are full. It is important to count the electrons to make sure they match the count you had at the beginning. If the totals don't match up, you may be leaving out a double bond somewhere.

Samurai raised his hand, finally ending Professor Oak's seemingly unending monologue. "I have a question I wish to ask. If you have a molecule such as Nitrous Oxide with several ways to draw a Lewis structure, how will you be able to tell what the correct one is?"

"Ah, very good question," Professor Oak replied. "One way you can do is to figure out the formal charge of each electron in the molecule. To find the formal charge, you would take the number of valence electrons, and then subtract the number of bonds it has and the number of unpaired electrons it has. Let's take Nitrous Oxide as an example. Nitrous Oxide has two Nitrogen atoms and an Oxygen atom. This totals out to sixteen valence electrons. Nitrogen is the least electronegative, so it will go in the middle. We'll draw a Nitrogen atom and an Oxygen atom around it, and give it single bonds. If we give it single bonds, then fulfilling the octet rule would require twenty electrons, which is far too many, so we'll give them double bonds. The central nitrogen has completed its octet, and the two atoms to the side need four more electrons to complete their octets. We are left with sixteen electrons, which is what we want. This is one Lewis structure. We can make another Lewis structure by giving Nitrogen and Nitrogen a triple bond, leaving a single bond for Oxygen and Nitrogen. This would require one of the electron pairs on the Nitrogen atom to go the Oxygen atom, but it fulfills the sixteen electron limit.

"Now we can find the formal charges. In the first Lewis structure, the central Nitrogen has five valence electrons, and it has four bonds. Five minus four is +1. This is its formal charge. The other Nitrogen five valence electrons, but it has two bonds and four unpaired electrons. Five minus two minus four is –1. It has a formal charge of -1. Finally, Oxygen has six valence electrons. It has two bonds and four unpaired electrons. Six minus four minus two is 0. Who can do the formal charges for the other Lewis structure?"

"I'll do it," said Richie. "The central Nitrogen still has four bonds, so its formal charge is still +1. This time, though, the other Nitrogen has three bonds and two unpaired electron. Five minus three minus two is 0, so its formal charge is 0. The Oxygen has six unpaired electrons and one bond, so its formal charge is six minus six minus one, or -1."

"Very good, Richie," said Professor Oak. "Another important thing to know about Lewis structures is that the sum of the formal charges must equal the overall charge of the molecule or the ion. Well, generally the Lewis structure with less formal charges is the correct one, but in this case, they both have equally low formal charges. Then, these two forms of the molecule can coexist, although the second one may be a little bit more favored because the negative formal charge is on the oxygen – the more electronegative element. We call these alternate forms resonance structures. The overall form of the molecule is somewhere in between these two structures.

"Finally, there are some intermolecular forces that are not quite as strong as bonds, but they can still affect the physical properties of a molecule, especially its boiling points. The first one is dipole-dipole interaction. Polar molecules usually have regions with a partial negative charge and a partial positive charge. Whenever two polar molecules are near each other, then the charges would arrange themselves to create an electrostatic interaction. This polar interaction would increase the boiling point when compared to non-polar molecules. The Hydrogen bond is a special polar interaction. Whenever Hydrogen is bound to a very electronegative atom that has lone electron pairs like Nitrogen, Oxygen, or Fluorine, then there is a very big partial positive charge on the Hydrogen, and it would be able to orient itself to other partial negative charges. Hydrogen bonds are important in water, as well as alcohols, amines, and carboxylic acids. These bonds also keep DNA together.

"A third intermolecular force is electrostatic interaction. This usually happens when ionic crystals are dissolved in a polar solvent like water. The partial charges would arrange themselves around the freed cations and anions. The final intermolecular force is the weakest individually, but it usually has the highest accumulated effect. This is known as dispersion forces, or London forces, or Van der Waals forces. Basically, electrons can spin in ways to create partial positive and negative charges even in non-polar molecules. These small charges can nevertheless lead to intermolecular forces. They may be small in small non-polar molecules, but once the molecules get to be a certain size, then the accumulated effects can be so great that they have higher boiling points than polar substances like water."

The bell rang. "Ah, looks like class is over. Good thing I got through everything I wanted to cover," said Professor Oak. "Be sure to do the homework, and read the chapter on Stoichiometry. It's not difficult math, but it's important that you get it right."

Brock, Misty, and Richie got up to leave, but noticed that Ash had fallen asleep on the desk. As they shook him awake, Gary started laughing. "Hah! I knew he doesn't have what it takes to last in this class! And he thinks he can beat me! What a joke!"

Ash was groggy as he woke up, but he still heard the comments Gary made. "Someday I'll show you, Gary!" he cried, and then fell back onto the desk to sleep some more.

--

AUTHOR'S NOTES: This was the chapter where I had a real problem with coming up with good examples, and it probably isn't the last. Professor Oak goes on a long, stirring monologue that runs for almost two pages, and a little bit later he embarks on a one-page uninterrupted lecture. I guess it's because the periodic table is so concrete it's hard to integrate something like Pokémon to it. The original story had pictures to end the monotony, but doesn't like pictures.


	7. Chapter 7: Stoichiometry & Thermodynamic

Brrring

DISCLAIMER: Nintendo owns all. I own nothing.

--

Ash Ketchum walked into the classroom with great bags under his eyes. He looked as though he hadn't slept in days, which he hasn't.

"Man, you look rough," said Richie. "What have you been doing?"

"Well, you did say that in order to beat Gary in this class, I'd have to become familiar with the material in this class, so I read the entire book in one night."

"Wow, that's impressive," said Tracey. "How much of it do you remember?"

Ash thought for a little bit. "You know," he said at last, "I don't think I remember any of it."

"That's the problem with cramming," said Misty. "You read things, but you don't remember any of it."

"It won't help you on any of the tests that way," said Brock. "You have to make sure that you understand everything about a topic before you move on."

"Then what do you want me to do?" cried Ash.

"You can make sure you understand everything about a topic before you move on," Brock repeated.

"It'll also help to pace yourself, and not read everything at once," said Richie.

"Ha! Ash is so stupid he doesn't even know how to study!" laughed Gary. Brock and Richie had to restrain Ash to prevent him from going after Gary.

"Maybe you should get some more sleep, Ash," said Misty, "but don't do it in class."

Team Rocket came rushing in before the bell rang. They smelt of burned flesh, and Jessie's hair was puffed out in a massive afro. "I knew we had to bring our rubber suits," cried Meowth.

"Oh, shut up," said Jessie. "What are you going to about with my hair?"

Professor Oak walked in, taking no notice of Team Rocket's condition. He set his books down and said to the class, "I realized that you may have found yesterday's lecture to be on the boring side. I apologize for that, but I hope that you still paid attention, because yesterday's material was very important. I will try to make the material more interesting today. Today, we will be talking about compounds, which are the products of the chemical bonds we discussed yesterday, and stoichiometry, which is the mathematics involving these compounds.

"There are two types of compounds based on the two types of bonds. Ionic bonds lead to ionic compounds, which are solid crystals made up of these ions. Covalent bonds lead to molecular compounds, better known as molecules. Ionic compounds are generally too large to have molecular weight. Rather, we measure them by their formula weight, which usually is just the atomic mass of one set of ions. Molecules have a definite shape, so we can use a molecular weight that is simply the sum of the atomic masses of the components. There is no mass defect related to molecules.

"The molecular weight of one molecule is measured in atomic mass units, but the amu is too small to be really meaningful in calculations, so it's best to convert it to a unit we can use. We do this by using the mole. A mole is equal to the number of molecules so that the mass in grams, known as the molar mass, is equal in magnitude to the atomic mass in amu. For example, there is one mole of molecules in 12 grams if Carbon-12. We talked about the number of molecules are in a mole. Who still remembers that number?"

A lot of students raised their hands, but Professor Oak picked Duplica to speak. "One mole is equal to 6.02 x 1023 molecules," she said.

"That is correct," said Professor Oak. "This number is also known as Avogadro's Number."

"I have a question," said A.J. "These molecules are too small to count, so how do chemists know that there are 6.02 x 1023 molecules in a mole?"

"Well, actually, the person that came up with this number was a physicist. He wanted to find the number of particles in a unit of gases, and his calculations involved the gas constant, which we'll discuss in about two classes, and the Boltzman constant, which we'll talk about in a future date. The math is really too complicated for our purposes, but you're welcomed to study it up yourself if you really want to know.

"There is another side to molar masses that we'll be coming back to again when we talk about acids and bases, but it's good to introduce it now. It is the idea of equivalent weight. The general definition of an acid is a proton donor. Many acids donate one proton, but some can donate two or three. These polyprotic acids can donate two moles of proton in one mole of product. The equivalent is the unit to measure the number of protons an acid can donate. The gram-equivalent weight is the molar mass of the acid divided by the equivalents, which can be protons for acids and hydroxyl groups in bases.

"The good thing about compounds is that any sample of a compound has the same mass ratio as another sample of a different compound. This is because molecules are always made up of the same atoms. The make-up of molecules is shown in molecular formulas. The molecular formula basically tells you how many of each atom is in a molecule. A molecular formula may be simplified like fractions to an empirical formula. So ethane has a molecular formula of C2H6, and an empirical formula of CH3. You may be wondering what good is an empirical formula…"

"That's exactly what I was wondering," Meowth interrupted. "It seems completely useless."

"Well, perhaps it seems useless, but empirical formulas are good for our first stoichiometric calculation: percent composition. The percent composition is the mass of a certain element in a formula divided by the overall molar mass, and multiplied by 100 to give a percentage. Ethane has a molar mass of 30, so the percent composition of Hydrogen is 6/30, or 20. However, if you try it with the empirical formula, you see that you'll get the same answer: 3/15, or 20. If you have a very large molecule, it's easier to use the empirical formula than working with the large numbers. You can also use empirical formulas in backwards calculations. If you're given the percent composition and an overall molar mass, it's possible to divide the percent compositions by the atomic mass to find the empirical formula, and then multiplying them by whole numbers until the mass of your formula matches the molar mass.

"The other stoichiometric calculations involve reactions between different molecules or ions. These reactions are at the heart of chemistry, so it is important to know them. The reactants are the compounds that undergo reactions, and the products are, well, the products of these reactions. The basic way to write out reactions is to have the reactants on the left and the products on the right, and you have an arrow pointing from the left to the right. There are four basic types of reactions. What are they, and what are some examples of each of them?"

"A combination reaction combines two reactants to form one product," said Joe. "One example is to take hydrogen gas and oxygen gas to make water."

"A decomposition reaction is the opposite of a combination reaction. It's when a reaction breaks down to form two products" said Todd. "One example is the decomposition of hydrogen peroxide H2O2 into water and oxygen."

"A single displacement reaction is a reaction where an ion in one compound moves to another compound," said Samurai. "A reaction is most redox reactions in electrochemistry."

"A double displacement reaction is also known as a metathesis reaction," said Lara. "Most neutralization reactions between acids and bases are metathesis reactions."

"Very good," said Professor Oak. "I'm glad all of you did your readings. These reactions will be the focus of the rest of this chemistry unit and the organic chemistry unit that follows. I would like to make a point about the redox reactions Samurai talked about. We'll talk more about these reactions in a later class, but these are reactions where ions react with a solid electrolyte. One example is the reaction between Zinc and Copper Sulfate to form Copper and Zinc Sulfate: Zn + CuSO4 - Cu + ZnSO4. Since there are ions involved, it is possible to write an ionic equation, which separates all of the ions: Zn + Cu2+ + SO42- - Cu + Zn2+ + SO42-. Redox reactions involve the transfer of electrons, so it is important to determine which species are actually involved in the electron transfer."

"Well," said Richie, "Sulfate always remains in its anionic form, so it's not involved. Copper and Zinc both change their ionic charges, so they're obviously involved."

"That's correct. Sulfate is what is known as a spectator ion, since it's not involved in the reaction. Copper and Zinc are involved, so you would write a net ionic reaction showing the change in charges: Zn + Cu2+ - Cu + Zn2+.

"The first important thing you need to do before doing any stoichiometric calculations is to balance an equation. The important thing to know about these reactions is that the elements are conserved, so you need the same number of one element on the left as on the right. Let's do a sample: the combination reaction between Hydrogen and Oxygen gas to form water: H2 + O2 - H2O."

Brock gave it a try. "There are two Hydrogen atoms and two Oxygen atoms on the left, while there are two Hydrogen atoms and only one Oxygen atom on the left. Well, you need to multiply the Water by two to get to two Oxygen atoms, but that gives four Hydrogen atoms. You need to multiply the Hydrogen gas by two to get to four Hydrogen atoms: 2H2 + O2 - 2H2O."

"That is correct," said Professor Oak. "Who can tell me exactly what this means?"

"For every two moles of Water that forms, you need two moles of Hydrogen gas and one mole of Oxygen gas," said Misty, very succinctly.

"Very good, Misty. Now that we have the balanced equation, we'd be able to do stoichiometric calculations. The most basic calculation is to calculate the expected yield of a reaction. In these reactions, you're given a mass of the reactants, and you should be able to tell us how the mass of the products you have. Let's say you have eight grams of Hydrogen gas and 32 grams of Oxygen gas. How many grams of Water should we expect to get?"

"That's easy," said Ash. "You should get eight grams of Water, because you have 32 grams of Oxygen, but only eight grams of Hydrogen, so Hydrogen is the limiting reagent, so TAKE THAT, Gary."

"It's good that you know about the limiting reagent," said Professor Oak, "but you made one common mistake that people make in these stoichiometric calculations. You based the limiting reagent on mass, but what you must do is first convert all of the reactants to moles. The molar mass of Hydrogen gas is two grams, and the molar mass of the Oxygen gas is 32 grams."

"So you obviously have four moles of Hydrogen gas, and one mole of Oxygen gas," Gary said snobbishly, "The molar ratio between Hydrogen and Oxygen and Water in this reaction is 2:1:2, so Oxygen is the limiting reagent in this case. Two moles of Hydrogen would react with one mole of Oxygen to form two moles of Water, so you'll get 36 grams of Water. So who's going to take it now, Ash?"

"Grr, that Gary!"

"That's correct, Gary," said Professor Oak, and Gary's cheerleaders cheered wildly. "It is important to convert everything into moles, and then figure out the limiting reagent based on the balanced equation to be able to find the expected yield. However, in a lab you'll never get the yield that you expect. The yield you do get is known as the actual yield. The percent ratio between the actual yield and the expected yield is known as the percent yield. So what is the percent yield in our reaction if you get 24 grams of Water?"

"The percent yield is 67, because 24/36 is 2/3," said Meowth.

"Aw, man! I KNEW that answer too," Ash said disgustedly.

"It's okay," said Professor Oak. "The important thing is that you knew how to do it. Well, that's all I have about stoichiometry. There are a few minutes left, so I can start talking about our next product, thermodynamics. Thermodynamics is the study of heat, and it connects chemistry with physics, since it incorporates energy, work, and chemical reactions. Our talk about heat has generally been calling it a waste energy due to friction, air resistance, or elastic collisions. However, it has many useful applications. For example, it is heat energy from the Sun that keeps the temperature on Earth within livable range. It is heat energy from the pots and pans that cook our food. And it is heat energy emitted by Charmander's flames that keeps us warm when we're stuck in an icy cave."

"How did Professor Oak know about that?" Ash wondered.

"Heat can have a unit of Joule, because it is a form of energy, but a more common unit is the calorie, which is equivalent to 4.184 J. You often hear about calories in food, and one calorie in food is actually a kilocalorie, which is calorie with a capital C. Heat can be transferred in one of three methods. Conduction is a direct transfer of heat from one material to another. That is why Magnemites become hot to the touch once they come in contact with a Charmander's flames. Convection is a transfer of heat as material rises once they heat up and sink as they cool. It happens in a boiling pot of water, in a room, and in the mantle of the Earth. Radiation is the transfer of heat through waves. This how heat from the Sun reaches the Earth.

"Whenever we talk about heat in this unit, we are talking about heat in a certain system, which is defined as the particular part of the universe that is being studied. So it can be a chemical reaction or the Magnemite that is being heated by the Charmander. There are three types of systems. An isolated system cannot exchange energy or matter with the environment outside of it. An example is a bomb calorimeter chemists use to measure heat exchange of a reaction. A closed system can exchange energy with the environment, but not matter. An open system is the Charmander and the Magnemite, as energy and matter are being exchanged with the environment.

"There are three Laws of Thermodynamics which describes thermodynamic processes. The First Law of Thermodynamics connects energy, heat, and work. The energy is the internal energy of a system, and includes all potential and kinetic energy of the molecules. The pressure is the force experienced per unit area. This can apply to liquids or gases. Work in thermodynamics is not exactly the force distance value we're so familiar with. Rather, it's the pressure times the change in volume. Work can be done ON the system and BY the system. Work done _on_ the system by the environment would lead to a decrease in volume, which would lead to negative work. Work done _by_ the system on the environment would lead to an increase in volume, which would lead to positive work. The first law states that the internal energy U is equal to the heat Q minus the work W. Sign convention for heat is that heat gained is positive heat and heat lost is negative heat.

"There are three types of systems. An isothermal system has no change in temperature. An isobaric process has no change in pressure. And an adiabatic process has no heat exchange. However, only the adiabatic process would have a major effect on energy."

"What would the work be if you don't have an isobaric process?" asked Brock.

"Well, you may be tempted to just say it's the change in pressure times the change in volume, but it's not quite that simple. It would help to draw a plot with pressure on the _y_ axis and volume on the _x_ axis. If you draw any changes in pressure or changes in volume, then the work would be the area of the region underneath the volume curve.

"The second law of thermodynamics is much simple than the first. It means that with every reaction, the disorder in the universe would stay the same or increase. It will never go down. This disorder is known as the entropy, and it plays an important role in chemical reactions. You can calculate the change in entropy through isothermal processes. The change in entropy is heat exchange divided by the temperature.

"The final law states that there is an absolute zero temperature when all molecular motion is stopped, and entropy would be at a minimum. It is impossible to reach this absolute zero, although researchers have come close. There is also a zeroth law that came before the other laws. It states that two systems put in contact with each other will exchange energy until they reach the same amount of heat energy per unit volume. This may seem like common sense, but it is still important to keep in mind.

"Well, that is all I have for you today. We'll continue our discussion of thermodynamics in our next class. See you then."

Misty and Brock stood up to get ready to leave, but they saw that Ash was asleep AND he was drooling on the table. Professor Oak watched them try to shake him awake. He knew that he'd have to spend a lot of individual time helping Ash understand everything, and that it would be a task more daunting than pushing a Snorlax on ground with a coefficient of static friction of 10. "It's going to be a long year," he said.

--

AUTHOR'S NOTES: Originally this chapter was only supposed to contain stoichiometry, but that section turned out to be far too short, so I went on and talked about the laws of thermodynamics, which was a shame because that seemed to be stuff that would be Blaine's specialty. I want to get as many of the gym leaders in as guest instructors as I can. In the end I let Blaine have thermochemistry, which he probably doesn't seem as closely associated with, but in the end it doesn't really matter.


	8. Chapter 8: Thermochemistry

Brrring

DISCLAIMER: All of the characters in this story are properties of Nintendo, but the close relationship between Blaine and Lara is something I made up…but a family like the Laramie's would have some connections, right?

--

Ash, Misty, and Brock walked into the classroom, and to their surprise the imposing figure was not Professor Oak but a tall, lanky man in a white lab coat with a bald, shiny pate.

"Hey, are you going to be our sub for today?" Brock asked.

"Yes and no," the man replied with a high-pitched, scratch voice. "Professor Oak is going to be around, but I'm going to be doing the lecture for him. I hope you're all red hot and ready to go!"

"Well, I'm red hot, but I can't say I'm ready to go," said Ash as he looked warily at Gary, who was basking in the love and admiration from his cheerleaders.

The bell rang, and Professor Oak walked into the classroom. He walked up to the man and they exchanged a few words. Afterwards he turned to the class and announced, "Well, class, we are honored to have a very special guest with us today. I've asked Blaine, the Cinnabar Island gym leader to take time out of his busy schedule to give a guest lecture about thermochemistry."

"Oh, Uncle Blaine!" Lara cried excitedly, "it sure has been a while! You never told daddy that you were going to be teaching this class."

"Well, it certainly came as surprise to me as well," Blaine said. "How's Rapidash been doing?"

"Wait, do you mean to say you're related to Blaine?" A.J. asked.

"Well, we're not related by blood," replied Lara, "but Blaine's a good friend of my daddy's, and he's been an invaluable help in teaching me how to raise my Rapidash."

"Well, it certainly is an honor to be in your presence," said Gary. "Of course, I don't think Ash deserves to be a part of this historical event."

"Oh yeah, Gary? I'll have you know that I've met Blaine more times than you have!" He pulled out his Volcano Badge and cried, "TAKE THAT!"

Gary only shook his head and sneered. "You do know that only a novice shows off his badges."

"Ash may not be the most proficient at these problems in science," said Samurai, "but he certainly is no novice."

"Well, even the bug catcher is defending Ash now? Man, it must be good to be the main character," said Gary.

"I hope you're all done with your bickering," said Blaine, "because I'm certainly rearing to go! As Professor Oak said, Blaine's my name and heat is my game! Professor Oak said that he's already gone over the basic laws of thermodynamics with you already, so I won't go over it with you again, but let me test what you all know. What are the three laws of thermodynamics?"

He pointed his finger at Todd, who answered, "The first Law of Thermodynamics states that the change in the internal energy of a system is equal to the net heat minus the work done by the object, which is the pressure (force over area) times the change in volume."

Blaine stopped him and pointed to Joe, who replied, "The second Law of Thermodynamics states that the total entropy in a universe would stay constant or increase. The entropy in an isothermal system is equal to the heat divided by the temperature."

Blaine went on and scanned the room to look for the next person to answer and was surprised to see Team Rocket huddled in the back after arriving late following another failed Pikachu kidnapping caper. He pointed at Meowth, who mouthed 'Me?' for confirmation before answering, "The third Law of Thermodynamics says that there's an absolute zero value where all motion of matter ceases. This value is equivalent to 0 K."

"That is correct," said Blaine. "It seems like old Okie here still has it in him." He turned his focus to Team Rocket. "You were the ones that destroyed my volcano arena and almost led to an eruption that could have killed thousands of innocent lives. I'll forgive you for that, but I do have to ask you one question. When you fired those ice missiles, none of the missiles actually came in contact with the rock, but why did the arena walls collapse anyways?"

"How the heck should I know," groused Jessie. "All we know is that we couldn't get our hands on your Magmar."

James and Meowth didn't know, so Misty came out with the answer. "Well, the ice missiles would lower the temperature of the rocks that make up the walls. When solids and liquids vary in temperature, then their volumes would vary as well. So when the temperature of the walls decreased, the wall began to fall apart because of the decreasing volume."

"Ding ding ding! Misty comes up with the correct answer once again. All right, I only wanted you to know the general physical concepts of this. Professor Oak will be the one that teaches you how to calculate the changes in volume. We'll be moving in a completely different direction now, that of the field of thermodynamics in chemistry. You probably already know from Professor Oak's lecture that enthalpy and entropy are important values in thermodynamics. Well, they're the important state functions in chemical reactions." He pointed at Tracey, "You! Tell me what a state function is."

Tracey frantically searched his mind for the correct answer. "A state function is a path-independent function whose value remains constant no matter what direction you take. Another example is work."

"Very good," said Blaine. "A chemical reaction is quite often not simply a sudden, one-step reaction between the reactants. There are several smaller steps that would sum up to equal the final equation. Now all of these smaller steps are individual reactions themselves, and they come with their own changes in enthalpy and entropy. There are many ways to find the total change in enthalpy of a reaction. I trust you should all have read the readings, so you should be able to tell me what these different ways are."

He spun around and pointed at Duplica. "The first way is by taking the sum of the standard heats of formations (ΔH0f) of each of the products and subtracting it by the sum of the ΔH0f of each of the reactants. The standard heat of formation is the enthalpy change in the formation of the molecule from its individual elements," she answered.

Blaine turned to Richie, who replied, "Another way if you have the ΔH0f of various reactions but not individual ΔH0f is to take the reactions and sum them up to equal the ΔH0 of the final reaction. This may require flipping a few reactions around or multiplying them by a whole number. Then you just sum the ΔH0f of all of these reactions. This is Hess's Law."

"Okay, very good," said Blaine. He turned and stared intently at Ash. "So, what are a few examples of some of the more basic types of ΔH0f for reactions?"

Ash hemmed and hawed and squirmed around nervously in his chair. He refused to give up because he knew the insults Gary would heave at him if he did. Richie passed him a note that said, "Remember what we went over about burning and boiling." Ash thought about it for a few moments, and his eyes lit up. "There's the heat of, um, boiling, and that is the conversion of a liquid molecule to a gaseous molecule. And then there's the heat of, um, burning, which is the, uh, burning of a substance to produce water and carbon dioxide."

"Gee, it sure took you long enough to get that," said Gary. "Makes us think that you didn't really know the answer at all."

"Of course I know the answer, Gary!" cried Ash. "I just said so, didn't I?"

"And rightly so," said Blaine. "Those are correct, except that the proper names are heat of vaporization and heat of combustion."

"I knew that," said Ash.

"All right," said Blaine as he turned to Gary. "What's the third way to find the ΔH0f of a reaction?"

Gary paused. He knew Hess's Law and the final minus initial, but couldn't remember the final one. He looked at his grandfather, but Professor Oak only looked back at him without any indication of giving him the answer. However, Meowth grew tired of waiting for Gary and began to speak, "It's the bond…"

"Bond dissociation energy!" cried Gary. "It takes heat to separate each individual bond in a molecule, and different bonds have different dissociation energies. If you take the sum of the energy needed to form all of the bonds in the products and subtract from it the sum of the energy needed to break all of the bonds in the reactants, you'll get a good estimation of the ΔH°f. It was just on the tip of my tongue. Thanks, Meowth for the reminder!" Gary's cheerleaders cheered for their hero, chanting "Gary, Gary, he's so neat! He knows all 'bout standard heat!"

Blaine quieted the cheerleaders and moved on. "That's right. Those are the three ways to find the ΔH°f of a reaction. The ΔH°f of a reaction can be either positive or negative. Who can explain what this means?" He looked around and pointed at Brock.

"If the ΔH0f is positive, then it's an endothermic reaction. This means that you need to put in a net heat for the reaction to proceed. If the ΔH0f is negative, then it's an exothermic reaction. This means that a net heat will be released in the course of the reaction."

A.J. raised his hand. "I probably should have asked this earlier, but what exactly is the difference between ΔG0 and ΔG?"

"Ah, the degree sign just shows that the reaction is done in the standard state. The standard state is the state where most reactions are done: 1 atm and 298K. Entropy is another important state function in thermochemistry. As that kid over there told us," he said referring to Joe, "the entropy is equal to the disorder of a system. And just like ΔH0f, there is a ΔS0 for each reaction that is equal to the sum of the ΔS0 of the products minus the sum of the ΔS0 of the reactants. And the total entropy in the universe can pretty much only stay the same or go up. There's only one incident where the total entropy stays the same." He turned and pointed at Samurai, "what is it?"

"The change in entropy in a standard state would stay the same if the system was in equilibrium," Samurai replied.

"That is correct," said Blaine. "Enthalpy and entropy are nice, but they alone don't determine the spontaneity of a reaction. The spontaneity is actually determined by the change in the free energy of the reaction system. This free energy is also known as the Gibbs Free Energy. Each reaction has a ΔG0 that determines the spontaneity of the reaction. If the ΔG0 is negative, then the reaction will be spontaneous. If the ΔG° is positive, then the reaction will not be spontaneous."

Joe raised his hand. "Um, how exactly does that work?"

"Ah, very good question. Who thinks they have what it takes to explain it?"

Tracey spoke up. "Well, if you think about it, the free energy of a molecule is like the potential energy of the molecule. As you remember from physics, the potential energy tells you how likely the system is going to do work. If you have a negative ΔG0, then the reaction will occur and you end up with less potential energy than you started with. It makes sense for it to be spontaneous. It's like if you drop a ball and gravity will pull it down. On the other hand, if you drop a ball, and it goes up, then its potential energy will increase. It's the same in a reaction with a positive ΔG0. This won't happen unless you give it some extra energy, which is why it's not spontaneous."

"That's a good explanation," said Blaine, "a little wordy, but good. The Gibbs free energy can be related to enthalpy and entropy in this manner: ΔG ΔH - TΔS. Now who can explain this equation?" He pointed at Jessie.

"I don't want to," said Jessie.

"No? But I know you know it, and we're going to sit here until you decide to spill it."

They entire class sat silently for a few minutes before Jessie finally gave in. "Fine, I'll tell you. The ΔG is related to the ΔH minus the temperature times the ΔS. Are you happy?"

"I want more information from you. In what situations would the reactions be spontaneous or non-spontaneous?"

Jessie was still adamant, and at last James spoke up for her. "The reaction will always be spontaneous if ΔH is negative and ΔS is positive. The reaction will always be non-spontaneous if ΔH is positive and ΔS is negative. If ΔH and ΔS are both positive, the reaction will be spontaneous if the temperature is high enough. If ΔH and Δ° are both negative, the reaction will be spontaneous if the temperature is low enough."

"That's very good," said Blaine as Jessie gave James a good beating for participating in class. "The standard Gibbs free energy can also be related to the concentration equilibrium with this equation, ΔG° -RTln(Keq). I know you haven't done anything with the concentration equilibrium, but it's important to keep this in mind when you do." After that, Blaine glided to the white board and drew a curve on the board:

"This is a free energy diagram," he explained. "On the _x_ axis is the reaction mechanism, which is how the reaction proceeds. On the _y_ axis is the free energy. Who can explain how to interpret this diagram?" Once again, he scanned the room and settled on Todd.

"Uh, the diagram shows how the free energy changes as the reaction proceeds. The flat line on the left signifies the energy of the reactants, and the flat line on the right signifies the energy of the products. The difference in height between the flat lines is the ΔH, while the hump at the top is the transition state. A reaction always has to undergo some high-energy transition state. This transition state has no bearing on the spontaneity of the reaction, but it is very important to the rate of the reaction."

"That's very good," said Blaine. "Well, I've gone over just about everything because Professor Oak took the meaty things about thermodynamics, but I did leave one thing out. Does anybody know what it is?"

The class was quiet. Some students had no clue, while others seemed to have an idea but were afraid to answer. Finally Misty spoke. "You still haven't talked about specific heat yet."

"And Misty comes up with the correct answer once again! Why don't you explain specific heat to us?"

"Specific heat (c) is a physical property of a compound. It's basically the amount of heat needed to raise the temperature of one kilogram of substance by one degree. So the total amount of heat needed to raise a certain amount of substance by a certain amount of degree is the mass times the specific heat times the change in temperature: mcΔt. I know this because water has an especially high specific heat."

"Thank you for the explanation, Misty," said Blaine. "Well, let me turn things over to Professor Oak again."

Professor Oak stepped up to the front. "Thank you Blaine," he said. "Let's all give him a warm round of applause." When the applause subsided, he continued. "We are coming close to our first test. The next class will be about phases and gases, while the class after that will be about fluids and solids. This will cover all of the material that will be on the first test. We'll have one more class with material on the second test, and the class after that will be the test. I hope you're all well prepared. See you at the next class."

The bell rang, and most of the students left the room, talking about the upcoming test and Blaine's eclectic lecture style. While Misty, Brock, and Richie went to talk to Blaine with Lara and A.J., Ash walked over to Gary. "So, Gary, I guess we were even today."

"You only wish, loser. I don't know what Richie wrote on his note to you, but I bet it was the entire answer. At least I was able to know all everything related to bond dissociation energy just by hearing one word. We're still light years ago." He proudly walked out of the room with his cheerleaders, leaving Ash fuming behind.

At last, Ash cried out, "Just you wait, Gary! Someday I'll get you!"

--

AUTHOR'S NOTES: As I said in the last story, thermodynamics would probably be Blaine's expertise, but since Professor Oak got that out of the way, I might as well let him have thermochemistry. I gave him his finger-pointing method of lecturing, forcing students to use their prior knowledge, as a way to differentiate him from Professor Oak's purely didactic. And Blaine is the quizmaster, so you'd expect him to do stuff like that, right? Jessie's defiance was something I added as a whim, because you know she'd do something like that. I originally had a free energy diagram in here, but it won't appear on this site…neither would any of the superscripts or subscripts. :(


	9. Chapter 9: Phases and Gases

Brrring

DISCLAIMER: Pokémon belongs to Nintendo, and the MCAT material was lifted from the Kaplan book…but the idea to put both together was mine…yey.

--

Everybody was talking about the upcoming test as Ash, Misty, and Brock walked into the room.

"I'm going to have to stay up all night studying for it," Joe complained.

"I hope that it's okay to use calculators," said Tracey.

Ash looked befuddled. "When are we going to have a test?" he asked.

The entire class looked at him. "Uh, don't you remember that Professor Oak said we'll be having the first test in three classes?" asked Richie.

"No, can't say I do," said Ash.

"Yeah, it's probably because you've been spending the entire time sleeping, loser," said Gary.

"Why do you always say stuff like that to me?" Ash cried.

"Well, you kind of walked right into that," Misty said.

Brock agreed. "Yeah, wouldn't it be better if you just left him alone?"

"I tried," Ash argued, "but I'm sure you haven't had the experience of somebody saying bad things about you behind your back."

"I don't talk trash about you behind your back," said Gary. "I'm not that type of a person. I just come right in front of you saying how much of a loser you are."

"He's right," said Joe. "I don't think I've ever heard Gary say anything about you whenever you're not around."

"That's not very comforting," Ash muttered.

"You're just jealous of my sheer awesomeness," said Gary.

The bell rang, and Professor Oak walked into the room. He was followed quickly by Team Rocket, who were barely crawling into the room from the scratches, burns, and shocks they received.

"What happened to you guys?" asked Tracey.

"We were ambushed by a bunch of Ash's Pokémon. Who knew his mother was so good at using Pokémon," James said wearily.

"Well, I hope this would dissuade you from attempting any more unlawful entry," said Brock.

"It probably won't," Ash muttered.

"All right, class, it's time for us to continue. This is our second to last lecture with material that we'll cover in the first test, but hopefully it would be relatively easy. We'll be learning about the three phases of matter. We know by now that all compounds are made up of atoms and molecules, and we should know that all molecules have some attractive forces. Well, the strength of the attractive forces determines the phase that a molecule is in. Now, what are the three phases?"

"The first phase is the gas phase," said Duplica. "Gaseous molecules move the fastest, they're the furthest apart, and they're usually invisible, although you can sometimes see gas molecules, like steam or water vapor on a cold morning."

"The liquid phase is the next phase," said Joe. "Liquid molecules move slower and are the liquids now visible so they have a definite volume, but their shape is defined by their containers."

"The final phase is the solid phase," said Todd. "Solids have a definite volume AND a definite shape, although the shape can still be modified through physical forces. Molecules in solids are still moving, but not in a way that would disrupt the attractive forces."

"That's very good," said Professor Oak. "There is a fourth phase known as plasma when atoms move so quickly electrons are stripped from the atoms, but that's beyond the scope of this course. Theoretically, all compounds should be able to exist in any one of the three phases, but only in certain temperature and pressure. Water is the only material that actually exists in all three states in physiological conditions."

Professor Oak walked to the white board and drew a graph on the board:

"Now class," he said, "this is known as a phase diagram, and this shows you the temperature and pressure standards required for each state of matter. On the horizontal axis is the temperature, and on the vertical axis is the pressure. On the left is solid, and on the right is gas. The liquid state exists in the middle. This is an approximation of the phase diagram of water. The lines show the temperature and pressure conditions where two different states exist in equilibrium. In other words, these are the moments when there would be a phase change from solid to liquid or liquid to gas and so on. We'll discuss this later. For now, just focus your attention on this point O in the middle. Does anybody know anything special about it?"

Lara raised her hand. "Well, that's the place where the edge of all three states meet, so I guess it would be where all three phases would exist at the same time."

"That is correct, Lara. This is known as the triple point. Water's triple point exists at about 4.5mmHg and .001°C. The normal atmospheric pressure is 760mmHg, so it's not very easy to spot the triple point. The other point you have to take note of is this critical point (A) over here. Once you get past the critical point, which is the point after which the liquid form becomes one with the gas phase. You don't see this often in normal conditions, but it is still an important thing to know.

"Finally, notice how in water, the barrier between solid and liquid slants to the left – it has a negative slope." Several students nodded, and Professor Oak continued. "This slanting is due to the fact that the solid form of water is less dense than the liquid form, so as the pressure increases, the temperature where both phases are in equilibrium would decrease.

"The phase changes are an important thing to get out of these phase diagrams. The important thing to note is that a phase change is a form of equilibrium, the moment when both phases would co-exist. We'll learn more about equilibrium at a later date, but it is important to keep in mind that this is an equilibrium state and that the ΔG is equal to 0. Each of the phase changes has a different name depending on the direction you're going. What are the names of some of these phase changes?"

Professor Oak was unwilling to point like Blaine, so he ended up hearing a bunch of students shout them out. At last he had to call on students.

He called on Joe to name one. "Freezing is when you go from a liquid state to a solid."

A.J. was able to name another. "Boiling, or vaporization, is when you go from a liquid to a gaseous state."

"Condensation is the phase transformation from a gas to a liquid," said Samurai.

"Melting is the change between a solid to a liquid," said Tracey.

"Sublimation is when a solid goes directly to a gas," answered Richie.

"All right, very good; there is one more phase transformation, when a solid comes out from a gas. Who knows what that is called?"

The class was mostly silent. At last, Misty said, "Deposition."

"Very good; now, three of these transformations are exothermic, while the other three are endothermic. Which ones are which?"

Brock replied, "Sublimation, Melting, and Vaporization are endothermic, because bonds are being broken and you need energy to do that. On the other hand, Deposition, Freezing, and Condensation are exothermic, because you're going from a state with lower energy."

"That's correct, Brock. The temperatures where these phase changes occur have specific names: the boiling point for vaporization and condensation, and the melting point or freezing point for melting and freezing. There are a few important things you have to keep in mind about these phase changes. The first is on a thermodynamics scale. As you remember from last class, the temperature would increase once you add heat to a substance. However, it is important to note that once you hit the boiling point or the freezing/melting point, then there is NO change in temperature. Instead, the temperature would stay at the boiling point or the freezing/melting point until ALL of the substance has changed phases. This time, rather than the specific heat, you would use a heat of transformation, which can be heat of vaporization if you're talking about vaporization, and the heat of fusion if you're talking about freezing and melting.

"The other thing to note is that these melting points and freezing are constant for a pure substance, but they can change in solutions. These changes are known as the colligative properties, and they depend only on the quantity of the solute. So these changes would be the same if you dissolve sugar or salt in water as long as you dissolve the same amount of each. The quantity is measured in molality, which is moles of the solute over kilograms of solvent. This different from molarity, which we will discuss later.

"The first colligative property is freezing point depression. Freezing happens when molecules link together into solid crystals, which is why freezing is also known as crystallization. However, if you have solutes, the solute molecules or ions get in the way, and you need a lower temperature for the links to form. The change in temperature is the molality of solute times Kf, which is constant but different for every solvent.

"The second colligative property is boiling point elevation. The calculation of this change in temperature is similar to freezing point depression, only the temperature change is an increase rather than a decrease. You multiply the molality of solute with a Kb, a proportationality constant of the solvent. However, the reason behind this elevation is different. Each liquid has a vapor pressure. A substance vaporizes when the vapor pressure is the same as the atmospheric pressure. Yet the addition of a solute would lower the vapor pressure. Since the vapor pressure of the liquid is lower, you need to increase the temperature a little bit more to get the molecule to vaporize.

"The change in vapor pressure can be calculated. It is the mole fraction of the solute times the original vapor pressure of the solvent. The mole fraction is a lot like percent composition of a molecule, only you're calculating moles of a substance over total moles rather than mass of an atom over total molecular weight. The mole fraction can never be greater than one, so the vapor pressure would also lower. This is also known as Raoult's Law. You can also calculate the pressures of the solvent and the solute by multiplying the mole fraction of the solvent or the solute by the original vapor pressure of what you're looking for."

Brock raised his hand. "If the change in vapor pressure depends on the mole fraction, wouldn't Raoult's Law also be a colligative property?"

"That is correct, Brock. Vapor pressure lowering is the third colligative property. The final colligative property is osmotic pressure, and it is related to osmosis. We will learn more about osmosis later in the course, but basically if we have two compartments – one with pure water and the other containing salt water – separated by a semi-permeable membrane that allows only water to pass, water will diffuse from the compartment with pure water to the compartment with salt water so that the concentration difference would decrease. The concentration will never be exactly the same because the side with pure water will never get any solute because the membrane is permeable to only water, but eventually the water in the salt water side would exert a pressure on the membrane that would stop the flow of water. This pressure is osmotic pressure, and it is calculated with the formula MRT: the molarity (moles of solute over liters of solvent) times the gas constant times the temperature in Kelvin.

"We'll be talking more about pressure by a liquid in the next class, but these are the basics on phases and phase changes. Are there any questions?" Professor Oak scanned the room. Nobody raised their hand. "Okay, then we'll move on to the first phase: the gas phase. Even though gas molecules move randomly, there are still a few laws that define them. Before we go into these laws, we have to define something known as STP. Who can tell me what STP is?"

Duplica raised her hand. "STP is standard temperature and pressure, and this is the condition that scientists use when talking about gases. STP has a pressure of 1 atm, or 760mmHg and a temperature of 0 C, or 273. This is different from the standard conditions Blaine introduced in the last class, because standard condition is 25 C, or 298K."

"That's very good," said Professor Oak. "It is important to know that in doing calculations involving standard conditions or STP that you must use temperatures in Kelvin. Another important assumption we make in gas calculation is that gases are ideal. There are usually molecular attractions between gas molecules and there gas molecules all occupy a certain volume. However, these attractions and size differences are usually small enough that we can assume that they don't exist; that the gases are ideal. This assumption usually holds through until the temperature becomes very low and the pressure becomes extremely high. Now that we have these definitions out of the way, we can introduce the gas laws. What are these gas laws?"

Richie started answering, "Well, the ideal gas law is…"

"Ah, it's good that you know the ideal gas law, Richie," Professor Oak interrupted, "but it's important to introduce the fundamental gas laws that build up to the ideal gas law. We'll let some of the other students introduce these gas laws, and then you can talk about the ideal gas law." Richie nodded.

Lara answered first. "Boyle's Law states that the pressure of a gas and the volume it occupies are inversely related, but it is equal to a constant. So if the pressure of a gas rises, the volume decreases, but the product of the pressure and volume are equal."

A.J. answered next. "Charles's Law state that the volume of a gas is directly related to the temperature, so as the temperature rises, the volume also rises. Like Boyle's Law, volume over temperature is also equal to a constant. This is a fraction because of the direct relationship."

And then Joe answered, "Avogadro's Law states that the volume is also directly related to the number of moles present. Like the other two laws, this is also equivalent to a constant."

"Okay, Richie, take it away," said Professor Oak.

Richie cleared his throat. "The ideal gas law puts Boyle's Law, Charles's Law, and Avogadro's Law together, and it states that the pressure of a gas times its volume is equal to the moles of gas times the temperature in Kelvin times a universal gas constant, which varies depending on whether you're using Joules or Latm. It's 8.3145 J/(Kmol) and .0821 Latm / (Kmol)."

"Very good, Richie, and all the rest of you," said Professor Oak. "There are many things you can do with the ideal gas law other than calculate one of the unknown variables. You can use it find the density of a gas. So you know that the ideal gas is PV nRT. Well, the number of moles is equal to the mass over the molar mass, so that gets you PV (m/MM)RT. Density is mass over volume, so if you do the simple algebra, you would divide both sides by volume, and then divide both sides by (RT/MM), which is the same as multiplying (MM/RT). That gets you PMM / RT. You don't necessarily have to memorize this equation, but it's important to know how to derive it. The other thing is that you can use it to find the molar mass of a gas. Sometimes you may just be given the volume of a gas at STP, but you can still use it to find the number of moles as long as you keep in mind that a mole of an ideal gas occupies 22.4 L."

"Hmm. Octillery is the Pokémon number 224!" cried Ash. He paused. "I don't know how that's going to help me remember it, though."

"Haha, stupid Ash," laughed Gary.

"That will be enough, Gary," said Professor Oak, "but let us continue. As you all know, air is a mixture of many different types of gases. Each individual gas would exert a certain partial pressure on the container. As long as the gases have no chemical interaction on each other, then the total pressure is just the sum of the partial pressure of each individual gas, so PT P1 + P2 … + Pn. If you are given the total pressure and the number of moles of each gas, then you can find the individual partial pressure. Does anybody know how?"

Todd raised his hand. "You can calculate the mole fraction as you explained it earlier in the class, and multiply it by the total pressure."

"So it's kind of similar to Raoult's Law," Brock observed.

"Yes, that's a very good observation, Brock. After all, in both situations you're dealing with pressures. However, it's important to note that both laws are independent. The law of partial pressure was derived by John Dalton about 60 years before Raoult figured out the colligative properties.

"We all know that gas molecules move, and as we learned earlier, any movement is involved with kinetic energy. So it is possible to figure out the kinetic energy of a gas. However, it is important to make some more assumptions in addition to the assumptions we made about ideal gases. We must assume that gas molecules are in constant motion, and that any collisions between gas molecules are elastic."

"So there won't be any losses in kinetic energy," said Tracey.

"Correct. The kinetic molecular theory postulates that the average kinetic energy is proportional to one and a half times the temperature of the gas and also to a Boltzmann constant _k_: 3/2_k_T. However, the actual speed of a gas molecule is impossible to define because there are so many molecules. Therefore, rather than thinking of gases in terms of kinetic energy, it's more helpful to think of it in terms of a value called root-mean-square speed: urms. This value is basically the speed as calculated from the average kinetic energy, and you calculate it by taking the square root of 3RT over the molar mass of the gas.

"Diffusion and effusion are two types of movements based on kinetic molecular theory. Diffusion is the movement of a gas down a concentration gradient, and effusion is the movement of a gas molecule in a compartment through a small hole into another compartment. Even though these movements are different, they are still related to the molar mass of the gas. Based on what we just went over, can anyone think what the relationship between the rate of diffusion or effusion and the molar mass would be?"

Duplica spoke up. "Well, the root-mean-square speed is √ (3RT/MM). The molar mass is on the bottom, so it should be inversely related to speed, in that the large the molar mass, the slower it would go. So the rates would be inversely related to molar mass.

"That is correct. The rate of diffusion between two gases is inversely related to their molar mass, actually the square root of their molar masses, so r1 / r2 √ (MM2/MM1). It may be a bit confusing to have r1 and MM2 on the numerator, but as long as you keep in mind that the rate is INVERSELY related to the molar masses, you should be okay.

"So all this time, we've been assuming that the gas molecules are ideal, that they have no intermolecular attraction or volume, but we all know that this is not true. As we know, gases would deviate from ideal behavior at high pressures and low temperatures. As pressure increases, then the space between gas molecules would increase, and then the size of the molecules would become a problem. The assumption that the gas molecules have no volume would fall apart. And as temperature decreases, the root-mean-square speed of the molecules would decrease, and then intermolecular attractions would become more prominent, so that assumption would no longer hold.

"There is a formula that would take into account these deviations: the Van der Waals equation of state. This equation states than rather having PV nRT, you have (P + n2A/V2) (V – nB) nRT. This equation takes into account the deviations caused by intermolecular forces and the changes volumes. Crucial to each of these values are the constants A and B, which are constants different for each gas that corrects for the intermolecular forces and the volumes respectively. As you may expect, A would be large for more polar gases and B would be larger for, well, larger molecules. You don't need to memorize the equation, but you will have to know how to use it and the differences between A and B.

"That's all for today. Tomorrow we'll go over important information about liquids and solids. We'll be dealing with density and pressure, and most importantly, forces. See you next time, and good luck studying for the test."

The bell rang, and Ash quickly hurried out of the room in order to avoid Gary. Meanwhile, A.J. and Lara had become quite friendly. As they walked out of the classroom, they discussed the difference in lecture styles between Blaine and Professor Oak.

"You know, I do think your uncle Blaine is a much better lecturer with all of his finger-pointing and stuff. It may be nerve-wracking, but it sure keeps us on our toes and forces us to understand the material," said A.J.

"Yeah," Lara agreed, "Professor Oak just stands there and yaps. He's a nice guy and he knows his stuff, but it gets to be kind of boring."

They both shook their heads, and said at the same time, "This is going to be a long year."

--

AUTHOR'S NOTES: I originally had a nice phase diagram for water, but that didn't appear in this document. Oh well…you've probably seen it before if you decided to read this story. And in Ash using Octillery to remember 22.4 L / mole, I did something like that, except my example was that Jim Bunning and Catfish Hunter were both in the Hall of Fame with 224 wins. Bert Blyleven is not with 287 wins. Something is terribly wrong with that picture. A.J. and Lara's relationship near the end is an inside joke with my sisters, but I figure why not put it in this story?


	10. Chapter 10: Fluids and Solids

Brrring

DISCLAIMER: Nintendo owns all things related to Pokémon in this story, even Misty. Poor Misty. :(

--

It was the last class featuring material that will be in the first test, and the students were eager to learn, at least most of them were.

"There's so much stuff! How is it ever possible to know everything before the test?" cried Ash.

"It's not really that much once you think about it," said Richie. "You don't actually have to memorize certain facts. You just have to know a small number of concepts that you can apply in many different situations."

"But what about all of those equations? I can't possibly remember them all!"

"Oh, I'm sure Professor Oak would give us a formula sheet that we can use on the test," said Brock.

But when Professor Oak walked into the room when the bell rang, Todd asked him about the possibility of a formula sheet. "No, there won't be a formula sheet," he said. "I will give some of the more complicated equations on the test like the elastic collision equation Misty derived, but I expect you to know the basic equations."

A groan spread through the entire class. "Can you at least tell us what the basic equations are?" asked A.J.

"Certainly, A.J.; you'd definitely have to know the kinematic equations and the force equation for Newton's second law of motion. You need to know the basic trigonometric functions, the equations for kinetic friction, maximum static friction, and centripetal motion, W Fdcos(theta) ΔE, U mgh, K ½mv2, P W/Δt, pmv, τ rFsin(theta), Δp FΔt, the energy of a photon is E h_f_, the nuclear binding energy mc2, Aufbau Principle is E _n_ + _l_, the formula for percent yield, U Q-W, Q mcΔt or mΔL, the formulas for the four colligative properties, PVnRT, the law for partial pressure, and the laws you will be learning today."

Most of the students quickly scribbled down the equations as Professor Oak read them out, but Team Rocket came rushing in at this time and missed out on the list. "What's going on?" asked James as they sat down, but Samurai told them to be quiet as he kept on writing.

When Professor Oak finished, he went right into the lesson. "Today we're continuing from where we left off yesterday about the property of different phases. We'll be talking about fluids – which is the physics behind liquids and gases – and solids. The first two concepts should be very familiar: density and pressure."

Duplica spoke up. "Density is mass / volume. For an ideal gas, it is equal to the pressure times the molar mass divided by RT, where R is the ideal gas constant and T is temperature in Kelvin."

Brock explained pressure. "Pressure is an important concept in thermodynamics and in the ideal gas law. It is equal to the force / area. Dalton's law of partial pressure states that the total pressure of a mixture of gases is the sum of the individual pressure of gases. To find the individual pressure of a gas, you would multiply the total pressure by the mole fraction of that gas (mole of the gas / total mole.) This is similar to Raoult's Law, which describes vapor pressure lowering in a mixture. The new vapor pressure of a solvent is equal to the pressure of the solvent times the mole fraction of the solvent in the solution."

"Very good," said Professor Oak. "The important thing to remember about density and pressure is that there are a multitude of units for them. Density can be described in kg / m3 or g / cm3, which is the same as g / mL. If the density of water is 1 g / cm3, how much would it be if you convert it to kg / m3?"

The students used their unit conversion skills to work the problem out. Misty finished first, but Tracey was the one to reply. "There are 1,000 kg in a g, which means 1 g / cm3 will be equal to .001 kg / cm3. Multiply that by 1 x106 since there are 1 x 106 cm3 in a m3, and you'll end up with 1,000 kg / m3."

"That is correct, Tracey. It is very important to be able to do these unit analyses. A convenient measure for density is known as the specific gravity, which is the density of an object over the density of water, 1000 kg / m3. Now, there are also three units for pressure, and one of them has two names. The first unit is atm, which is the unit we use in the ideal gas law. This is equivalent to 1.013 x105 Pascals, our second unit. A Pascal is equal to N / m2, which we expect to get when we take the unit of the pressure equation. It is important to convert all of the pressures to Pascals when you are working with the Newton. The _third_ unit is the torr, which is also known as mmHg. In the distant past, people used to measure air pressure with the barometer, which measures the pressure based on the height of a bar of liquid mercury. 760 mg is known as the standard.

"Whenever you are submerged in a fluid, the absolute pressure on you would be P0 + ρgh, where P0 is the pressure at the surface of the fluid, ρ is the density of the fluid, g is the acceleration due to gravity, and h is your depth. The most common usage of this is our everyday life is with air. Air is a fluid, so we are constantly surrounded by a fluid. The air around us would apply a pressure on us, the atmosphere pressure, or Patm. Another important type of pressure is the gauge pressure, which is simply the absolute pressure minus the atmosphere pressure: P – Patm. Under normal circumstances, where Patm is the surface pressure, the gauge pressure is simply ρgh."

"There are two important physical principles related to density and pressure. The first is Pascal's Principle. This states that any change in pressure applied to a fluid in an enclosed space would be transmitted evenly to the rest of the fluid and also the container itself."

"Of what use would this principle be?" asked Gary.

"Sure, it may not seem like there'd be many applications for this, but it can be very useful. This is found mostly in hydraulic lifts used by mechanics in fixing cars or by Nurse Joys in helping large Pokémon. In a hydraulic lift, you'd have a U-shaped container with a small area on one side and a large area on another. What do you think would be the reason for this?"

"Well, the principle says that pressure is the same throughout," said Todd, "so if you apply pressure on one side, then there would be the same pressure on the other side, so F1/A1 F2/A2. If you have one side with a small area and a side with a large area, then you can get a large force on the large side by applying only a small force on the other side."

Duplica tried to make sense of this by putting it in an actual situation. "So if you want to accelerate a 460kg Snorlax 1m/s2, you'd need a force of 5060 N to overcome the downward acceleration due to gravity. Instead of getting under the Snorlax and applying a 5060 N force, you can put it on a hydraulic lift with an area of 10 m2, and if the other side has an area of .1 m2, you'd only need to apply a force of 50.6 N."

"Very good, Duplica. Now do you see the benefits of Pascal's Principle?"

Gary nodded. "But I know somebody who doesn't," he added.

Before Ash could say anything, Professor Oak went on. "The one issue with Pascal's Principle and hydraulic lifts is that not only is the pressure constant, but the volume of the area moved is also constant. The volume would be the area times the distance moved. Can anybody see the problem with this?"

Lara replied, "Well, going by A1d1 A2d2, if the side with the Snorlax moved up 1 m, then the smaller would have to go down 100 m just to keep the volume constant."

"Yes, so you'd have to keep this in mind if you ever want to design hydraulic lifts for use in Pokémon centers. Work performed by the lift is another constant in these hydraulic lifts."

"So the force multiplied by the distance traveled on one side would be equal to the force multiplied by the distance traveled on the other side," said Samurai.

"That is correct. The other important concept is Archimedes' Principle, which deals with buoyancy."

"Is Archimedes a Pokémon trainer?" asked Ash.

"Unfortunately not, Ash," said Professor Oak as Gary laughed at Ash's ignorance. "Archimedes was a famous Greek philosopher who was entrusted with the task of deciding whether or not the king's new crown was truly made out of solid gold. At that time, the density of gold was known, but there was no way to measure density without finding the volume, and the king really liked his new crown. One day, as Archimedes got in a bathtub, he noticed that water was displaced and spilled out over the rim. That got him thinking, and he came up with the set of equations that you will learn about now."

"Is it truth that he ran around naked through the streets crying, 'Eureka!'?" asked Tracey.

"Ew, we don't need that image in our head," said Duplica.

"We don't know that for sure. In fact, the entire tale is probably a legend, but his hydrostatic findings still stand today. When we put an object into water, there will be a buoyant force equal to the weight of the water displaced, which we is also equal to ρwgVw, where the ρ is the density of water, g is the acceleration due to gravity, and V is the volume of water displaced. This buoyant force is opposite to the weight of the object, which is mg, or ρogVo, where ρo is the object's density and Vo is the total volume of the object. If the object is floating, then the weight of the displaced water – or buoyant force – is equal or greater than or equal to the weight of the object. In this case, the volume of the displaced water is equal to the volume of the object _that is submerged_. If the object sinks, then the buoyant force is less than the weight of the object. This is why things less dense than water would float, because then ρo is less than ρw, and for the object to sink it has to displace a greater volume of water than the volume of the object. That's obviously not going to happen because ρo is already smaller than ρw."

"Is buoyancy the reason why a Pokémon like Surskit can walk on water?" asked Joe.

"Actually, a Surskit walks on water due to surface tension, which is based on intermolecular forces within the liquid. There are two major forces that are important: adhesion – the force between liquid molecules and the molecules of another substance – and cohesion – the force between liquid molecules themselves. Adhesion is why a Pokémon like Sunflora can absorb moisture from the ground, because the water tricks up through its stems. Cohesion is what leads to surface tension, and it is the most prominent on the surface of the liquid. The forces cancel out beneath the surface, but at the surface there is no opposing force on the outside, so there is a net force inwards. This force is constant throughout the surface, so it creates a tension like tension on a rope. This tension would be strong enough to support something like a Surskit as long as the contact area is small enough.

"A similar concept that applies to all fluids is viscosity. Viscosity is the measure of the internal frictional force within a fluid. So it would take more force to accelerate through a viscous fluid. Air is not viscous at all, so it takes almost no force to accelerate through air. However, it takes more force to accelerate through water because of the viscosity of water due to the Hydrogen bonding. And even though the sludge that makes up a Muk is a fluid, good luck trying to walk through a Muk. Believe me, I know.

"The viscosity of a liquid affects its flow. There are two major types of flows: laminar flow and turbulent flow. Laminar flow is the simplest type of flow, and it features layers of liquid sliding over each other. Once the velocity of the liquid passes a critical velocity, then it becomes very complex, with waves and eddies. This is turbulent flow. The critical velocity is dependent on the viscosity of the liquid. It's equal to NRη / ρD, where NR is a constant called the Reynolds number, η is the viscosity, ρ is the density of the fluid, and D is the diameter of the pipe. You don't have to memorize this equation, but you would have to understand the relationships that determine the critical velocity.

"Most of what we will study about fluid flow is related to laminar flow. The most basic idea is that a fluid's flow – the velocity times the area of the pipe it flows through – so the fluid's velocity is inversely related to the diameter of the pipe. We can put this together with what we learned about pressure to get Bernoulli's equation, which is like the conservation of energy equation for fluids. The equation is: P1 + 1/2 ρv12 + ρgy1 P2 + 1/2 ρv22 + ρgy2.

"You should know what all of the values are in this equation, but notice how there are two values that are very similar to the equations for kinetic and potential energy. These are precisely those energies, only for fluids. Based on this, you should be able to tell that velocity and pressure are inversely related at the same height. This concept is important in showing how bird Pokémon like Pidgeys and Fearows get their lift. Can anybody tell why?"

The students thought about it, and Misty answered, "Well, the wings of most bird Pokémon are designed so that the air flow above the wings would move quicker than the air flow below the wings. This means the pressure above the wings would be lower. There would be an increased pressure below the wings, and since force and pressure are directly proportional, there would be a force pushing upwards. This is also how tornados can lift the roof off of buildings."

"That's very good, Misty. When you do problems involving this equation, it is very important to draw a picture and be able to tell what values are at 0. This would make your life much easier.

"Besides these properties of fluids, there are important thermodynamic and elastic properties of liquids and solids that we must understand. The major thermodynamic property is thermal expansion, when a liquid or a solid expands or shrinks in response to changes in temperature. For example, a liquid would undergo volume expansion. If you heat up a liquid, its volume would increase, and vice versa. This is why people say it's better to fill up your gas tank in the cold. The change in volume is equivalent to the original volume times the change in temperature times a coefficient of volume expansion: ΔV VΔTβ.

"A solid can also increase with a change in temperature, but this expansion is usually linear – along one direction. The equation for this is similar in form to the equation for volume expansion, only with drastically different variables: ΔL LΔTα.

"Elastic properties are changes to a solid due to stress, which is exactly the same as pressure – F/A. The application of a stress upon a solid would lead to a strain, which are the physical changes due to the stress. There are three different elastic changes, and they all have an associated modulus that defines the likelihood of the solid to undergo that particular change. Let's just take the first elastic property as an example: the linear stretching of a solid due to the stress. So this is similar to the linear expansion of the solid in that it involves a change in length, but the change is due to a stress rather than a change in temperature. If you know the stress you apply to the object – the force applied over the linear surface area of the object – the strain would be equal to the change in the length of the object times the Young's modulus divided by the original length: F/A YΔL / L. You can rearrange the equation based on what you know and what you want to find. If you know Young's modulus and want to find the change in length, the equation is ΔL F/A / Y/L. If you know the change in length and want to find the Young's modulus, it would be Y F/A / ΔL/L.

"So what exactly does the Young's modulus mean?" asked A.J.

"I already explained it once, but it's worth going through again. The Young's modulus basically tells you the solid's resistance to stress. A material with a small Young's modulus, like the rope Jessie and James used to lift the Snorlax, would stretch very easily. However, something like a Skarmony's tail feather has a large Young's modulus, and stretches very little in the presence of stress. Yield strength is a value related to the Young's modulus, and it is the point of stress after which the solid becomes permanently deformed. You don't have to calculate it mathematically, but you do have to understand it.

"The other two elastic properties are very similar. The first is shearing. In shearing, instead of applying the stress directly, the stress is applied parallel to the surface and causes a deformation in a certain direction. A good example to use in visualizing shearing is a deck of cards:

"The equation for shearing is F/A Sx / h, where S is the Shear modulus, the x is the deformation distance of the object due to the stress, and h is the height of the object. The final elastic property relates to changes in volume due to the stress. Since this volume change can apply to fluids as well as solids, the stress can be expressed in ΔP as well as F/A. The equation is ΔP BΔV / V. As you might expect, B is the modulus. This time it is known as the Bulk modulus.

"Well, that's about the time we have for today, which works well because this is all the material we will cover for the test two classes from now. I will quickly go over the format of the test so we can jump right into the material in the next class. The test will be worth 100 points. There will be 60 multiple choice questions. Some of the multiple choice questions will be discrete questions, but most of them will be based on a passage that describes the situation in detail. The other 40 points will be from eight free response problems worth five points each that has you work out the problems on your own. You would be able to get partial credit even if you get the wrong answer as long as you set up the problem correctly. I hope this would clear up whatever anxieties you have. See you all next class."

The bell rang, and Gary was still laughing about Ash's question about Archimedes being a Pokémon trainer. "Man, I knew you were stupid, but I didn't know you'd think of a question like that!"

"Just ignore him," said Richie, but Ash was ready to rush up to Gary and fight him. It took the combined effort of Brock and Tracey to hold him back.

"Peh. Why would you hold him back?" said Gary. "A real gentleman fights only with words. This merely proves that Ash is nothing more than a rogue."

"Well, you're not one either!" shouted Ash.

"Yeah, whatever. Nothing you say is meaningful at all. Come on, ladies. Let's leave this ruffian with his despicable friends."

"Despicable?" Misty cried indignantly.

"You can't get away with calling my friends that!" cried Ash. "Someday I'll get the best of you, Gary!"

--

AUTHOR'S NOTES: Gary's sure got guts. He's gone from insulting Ash to insulting his friends as well. All this interaction between Ash and Gary were things I came up with off the top of my head. The real focus of this chapter was about fluid and solids, which I was never good at. I mean, I understand the equations and all, but I panic whenever trying to do math related to it. I also had a good picture documenting shear stress, but that didn't show up either. I could explain it in words, but it wouldn't be as effective as looking at a picture, so go look it up in a picture or something. Also, the structure of the test is a mix between the MCAT (passages with related questions) and a typical high school physics test (free response questions.)


	11. Chapter 11: Periodic Motion and Waves

Brrring

DISCLAIMER: Nintendo owns Pokémon, and almost all of the other video game paraphernalia that I own. Shogakukan / Viz Media own the Pokémon DVD that I watch, but Nintendo gets a cut of their profits too. Kaplan owns the book that I use, but I've been paraphrasing.

--

It was the last class before the first test, and most of the students were growing despondent.

"Why do we have to have a lecture about material that won't be on the test before the test?" asked A.J.

"I don't know, but you're right. There's should be a review session or something," said Lara.

Ash plodded into the room behind Misty and Brock. He was not looking forward to any of the classes anymore. He didn't understand the information, and he had to put up with Gary every time. Misty, Brock, and Richie went to talk to Professor Oak about Gary's behavior after the last class. Professor Oak said he'd talk to his grandson, but it didn't seem like any good would come from it. Gary smirked at Ash as he walked by.

Team Rocket had also been making visits to his house every day before class, presumably to get their hands on Pikachu. His other Pokémon have traditionally been successful in chasing them away, but each incident has been doing a certain level of damage to the house, and the costs are adding up.

The bell rang, and Team Rocket came scurrying into the room after their latest failure. Ash could tell that they've been trying every day. Their clothes are usually charred, and their skin is fried to a crisp. They do get points for tenacity, but why does it have to be with his Pikachu?

"Good morning, class," said Professor Oak. "Some of you have expressed concerns about us having a lesson the day before the test. Well, I hear your concerns, and I will be holding a review session in this room tonight at 6:00 pm. If you need help, bring your questions and I will be happy to answer them. We'll also be doing some sample problems. But we do have a lot to cover today, so we have no choice but to press forward.

"Our next few classes will be about various waves, so it's important to get the basic concept down. We'll start with a discussion of harmonic motion, which includes our prior discussion on forces and acceleration, but also introduces new topics like frequency and period. Then we'll go into waves, especially the two most prominent types of waves, sound and light.

"Waves and simple harmonic motion all fit into the same category of movement, which is known as oscillations. Oscillating systems are defined by repetition. However, there is symmetry to this repetitious movement that we can study. Simple harmonic motion, also known as periodic motion, is defined by an object going back and forth due to a restoring force across an equilibrium position. What are the two major examples of simple harmonic motion?"

Joe answered, "Springs and pendulums."

"That is correct. Springs and pendulums may seem quite different, but they actually have very similar properties. We'll start by investigating a spring. As you know, if you pull down on a spring and let go, it will keep going up and down until internal friction causes it to stop. This is an example of oscillations. The restoring force that drove it up and sends it back down is equivalent to –_k_x, where _k_ is a spring constant and x is the distance you pulled the spring down, and the negative sign means that the force is always pointed towards the equilibrium position when the spring is going away from that position. The spring constant is like the Young's modulus from the last class, and it defines how much a spring resists the force. A large _k_ would be hard to displace.

"There is another important value to a spring: the frequency. The frequency is a measure of how many times a spring makes a complete oscillation in a certain period of time, like a second. It has the unit of Hertz. The angular frequency (ω) is another frequency term, and it is a measure of the rotational frequency of an object, similar to the angular momentum we covered earlier in the class. The angular frequency of the mass spring system is equivalent to √ (_k_ / m), where m is the mass of an object attached to the spring. Knowing this definition, we can use it to find the acceleration of the system. Can anybody do it?"

Duplica answered, "Well, if F -_k­_x, that means ma -_k_x. Dividing the mass from both sides get you a ­ (-_k_/m) x. If ω sqrt(_k_ / m), then acceleration would simply be -ω2x."

"A very good derivation, Duplica," said Professor Oak. The ω is related to the frequency _f_ with the equation ω 2π_f_. This may make sense if you keep in mind that angular frequency measures the rotational frequency. A rotation is shaped like a circle, and what is the circumference of a circle?"

"2πr," Lara answered.

"Yes, so ω is 2π_f_. There is another important value, the period (T). This is basically a measure of the time it takes an oscillation to complete itself. It is the reciprocal – the opposite – of frequency, so as you can expect that it would be equal to 1 / _f_. Knowing the equation for ω, can anybody find the equation for T?"

"It's fairly simple," said Todd. ω 2π_f_. Ergo, _f_ ω / 2π, or (1/2π) sqrt(_k_ / m). The reciprocal of that would be 2π sqrt(m / _k_)."

"Very good, Todd. You can also find the kinetic and potential energies of the mass spring system. The kinetic energy is simply ½ mv2, which we should all be familiar with. The potential energy is slightly different, but it should be an easy equation to remember: ½ kx2. Can anybody think of when the kinetic and potential energies are at a maximum?"

Joe spoke up. "The kinetic energy is at a maximum when the system is going at its fastest, which is also when there is no potential energy. This should happen when the system goes back to its equilibrium point. The potential energy is max at each of the two ends, because the restoring force is pulling the system back."

"That is correct. This is all you really have to know about the mass-spring system. The pendulum is similar to this, except rather than having a force constant _k_, the force constant is mg/L. This should be enough information for you to derive all of the necessary equations."

"Well, in a pendulum, the displacement is equal to a certain angle θ, so F (mg/L) θ," said Duplica, "so ma (mg/L) θ and a (g/L) θ. This means ω √ (g / L)."

Tracey took over from there. "If ω sqrt(g / L) and ω 2π_f_, then _f_ (1/2π) sqrt(g / L), and T 2π sqrt(L / g)."

"The kinetic energy is equal to 1/2 mv2," said Brock, "and it is the maximum at the bottom of its swing. The potential energy should be mgh, and it's the maximum at the peaks of the swing."

"That's very good, class," said Professor Oak. "It seems like most of you have grasped the basics of simple harmonic motion. One thing you do have to keep in mind that a complete cycle is not complete until it goes back to its original position. So for a pendulum, it has to go back and forth before it can count as a complete cycle. Now we can move onto waves.

"Waves are similar to simple harmonic motion. In fact, if you connect a pen to a mass-spring system and have it draw a line as it moves up and down, you'd get one of the two types of waves, so it's obvious that frequency and period would be important in waves. As I said, there are two types of waves. The transverse wave is the wave that you get in a mass-spring system drawing. A transverse wave goes up and down perpendicular to the direction of propagation. Another example of a transverse wave is any wave you create by flicking a rope. The other type of wave is the longitudinal wave. A longitudinal wave is characterized by a change in pressure parallel to the direction of propagation. Either way, there are a few important equations that define them.

"The force constant _k_ is equal to 2π / λ, where λ equals the wavelength with a unit of meters, or the distance between the start and finish of a complete cycle. The angular frequency is equal to 2π / T, where T is the period. This basically goes to show you that _f_ really is 1/T, because ω 2π / T and also 2π_f_. Finally, there is an equation that you must memorize. Even if you forget all of the other equations, you should at least remember this one. The wave speed _v_ _f_λ. It makes sense if you think about it. Translational velocity displacement / time, so it's only natural that the wave speed is λ – equivalent to the displacement – times _f_, or 1/T.

"Another important variable that may not factor that much into equations is the maximum displacement of the wave, also known as the amplitude. If you take a look at this transverse wave, the amplitude is the distance from the middle to top of the wave, not the distance between the two wave peaks. This is important to remember."

Ash grimaced, waiting for the inevitable disparaging remark from Gary, but Gary just sat there with a pompous smirk on his face.

"The amplitude is important when you're talking about phase differences, the measure of how much two different waves are offset from each other. This is important when they meet. At the meeting, what you would do is add the amplitudes together. If two waves have a phase difference of 0° or 360°, then they are considered in phase. The peaks and the troughs would be at the sample place, and the final amplitude would merely be the sum of the other two amplitudes. However, if two waves have a phase difference of 180° or 540°, then the peak of one wave would coincide with the trough of another. If the two waves had the same amplitude, then they cancel each other out.

"An important concept related to phase difference that you need to understand is that of resonance. Everything has a natural frequency, the frequency where they oscillate without any forces. Now, as you know, a force would cause anything to start vibrating. If you want to keep it vibrating, then you have to apply the force periodically. If the frequency of the wave is out of phase with the natural frequency of the item, then the amplitude of the oscillation would be very large. However, the closer you get to the natural frequency of the object, the larger the amplitude would be. A good visualization would be pushing somebody on a swing. If you want the person to go higher, then you'd have to time the force so that it happens when the person is swinging back – when the potential energy is at a maximum. If you push when the person is swinging forward, then the effect of the force would be smaller, and you may even get hit in the face. Another example of resonance is how an Arbok's Screech can shatter a wine glass. The screech is a very high-pitched sound, so it has a very high frequency – close to the natural frequency of the wine glass. As you may expect, the amplitude of the wine glass's vibration would increase. It would increase until the amplitude is much greater than the glass's structure can withstand, and the glass would shatter.

"This seems like a nice transition to our next subject for today, the physics of sound. As you may already know, sound waves are essentially just variations in pressure caused by a displacement. We'll discuss how these sound waves are transmitted later in the class. It's good enough to know just what sound waves are for now. Sound waves can travel as long as there are material for the waves to travel through, like solids, liquids, and gases. The only thing it can't travel in is through a vacuum, because there is no matter to carry the sound waves. That's why there's the saying 'In space, no one can hear you scream.'

"Frequency and wavelength are important variables when it comes to sound, because as I had said the pitch of a sound – how high or low it is – depends on them. Pitch is important is important when it comes to the Doppler Effect, which describes the change in perceived frequency of an object in motion. Let's suppose you're sitting on a bench along the side of the road, when all of a sudden a Dodrio comes running by you. The Dodrio is going to be pretty loud as it rushes by you. Here's a recording similar to what you may hear."

Professor Oak stopped and played a tape on a tape player he had on its desk. The students heard a screeching Dodrio. The Dodrio started out far away, but as it got closer, its screeches seemed to increase in pitch until it passed by the recorder, after which the pitch began to decrease.

"If we assume that the Dodrio is going at a constant velocity, then it would be possible to calculate the new frequency. The new frequency _f_' is equal to _f_ (_v_ / _v_ -+ Vs), where _f_ is the emitted frequency, _v_ is the velocity of sound – 343 m/s – and Vs is the speed at which the source of the frequency is traveling. If you notice, there are both a subtraction and an addition sign. Can anybody guess what might be the meaning of these?"

Duplica raised her hand. "Well, if the Dodrio is traveling towards us, then the perceived frequency would increase, which means you need to multiply the emitted frequency by a value greater than one. You can only get that if you subtract Vs from _v_ to get a larger denominator. Similarly, once the Dodrio is traveling away from us, the perceived frequency would decrease. You'll need to multiply _f_ by a value smaller than one, which you get by adding Vs to _v_ to get a smaller denominator."

"That is correct. You can get the same effect if you're running by a grunting Slaking. As you run towards the Slaking, the pitch of the grunting would increase, and the pitch would decrease as you run by the Slaking. The effect is similar, but the calculations are slightly different. If you the observer is doing the running, then the equation becomes _f'_ _f_ (_v_ /- VO / _v_), where VO is the velocity of you the observer. As Duplica said, the frequency increases as you run towards the source – the Slaking, so you'll need to add your velocity to the speed of sound to get a larger numerator. On the other hand, when you're running away from the Slaking, the pitch would decrease, and you need to subtract your velocity to the speed of sound to get a smaller numerator."

"Now why the heck would you have two different forms of the equation if the concept is the same?" Gary demanded.

"Now why indeed," mused Professor Oak. "Ash, do you have an idea?"

Ash froze. He knew Professor Oak must have a reason of calling on him, but he couldn't think of anything. After wracking his brain for a while, he decided to make something up. "Uh…it's because…um…it's possible that…uh…you have both situations happening at the same time?"

Gary seemed unimpressed by Ash's answer, but Professor Oak said, "That's correct, Ash! It is possible for both situations to happen at the same time. So let's say you're jogging in the morning, and then a Dodrio comes rushing by and passes you."

"I'd stop and capture it!" cried Ash, boosted by his confidence.

"Perhaps it would serve you better to keep running and listen to physics at work. Before it passes you, you're essentially running away from the Dodrio while the Dodrio is running towards you. After the Dodrio left you in its dust, you'll be chasing the Dodrio who is leaving you further and further behind. What would be the equation for this?"

"Well, in the first situation, you're running away and the Dodrio is running towards you, so the equation is _f_' _f_ (_v_ – VO / _v_ – VS). VS would be larger, so you'll have a smaller denominator and so the perceived frequency would still be higher. Once the Dodrio passes you, it becomes _f'_ _f_ (_v_ + VO / _v_ + VS) since you're now chasing the Dodrio. The Dodrio's speed is greater, so you'll have a smaller denominator and the perceived frequency would still be lower. Oh, and a Rapidash runs must faster than a Dodrio."

"That it is, Lara, but Dodrio are more common around here. So I hope everybody now understands the Doppler Effect. Another characteristic of sound is its Intensity, which is the change in energy due to the wave per second – the power – divided by the surface area of something like your airdrum: I P/A. The intensity has a unit of W / m2, and it is the basis for the sound level, also known as the decibel scale.

"You are probably all familiar with the decibel, but it is quite complicated, as it is our first excursion with a logarithmic function, but it won't be our last. A logarithmic function is a function where one variable increases by a factor of 10. The logarithm of a number in a certain base is the exponent that the number must be raised to when put into scientific notation. For example, the logarithm of 1000 in base 10 would be 3, since 1000 1 x number with a logarithm of 4 would be 10000 – ten times greater than the number with a logarithm of 3 – because it's 1 x 104. Intensity is one such variable.

"The sound level β is equal to 10 log (I / I0), where I0 is the lowest intensity detectable by human ears – 1 x 10-12. The "10 " value is important, because that is where how we get decibels into double digits. Let us use the Whismur, Loudred, and Exploud as an example. A Whismur normally emits sounds with a sound level of 10 dB, but its cry can be as loud as 130 dB. A Loudred's cry can be as high as 140 dB, and an Exploud's cry can reach 170 dB."

"Bah, a Loudred's cry is only 10 times louder than a Whismur? That's kinda wimpy if you ask me," said Gary.

"Oh? Well, let's see how loud these sounds really are. What are the intensities of these cries?"

Brock did the math. "A Whismur's normal cry has an intensity of 1 x 10-11 W / m2, since that's only 10 times greater than the threshold of hearing. When it's threatened, its cry becomes 1 x 101 – 10 W / m2, and that is 1 x 1013 – ten trillion times louder than the threshold of hearing. A Loudred's cry is 100 W / m2, and an Exploud's cry is 100,000 W / m2. This is a full 10 quadrillion times louder than the threshold of hearing."

"That's right, Brock. And when you're dealing with Loudreds and Explouds, keep in mind that 160 dB is generally seen as the threshold where your eardrums rupture. So you must be careful when training these Pokémon."

"The final concept is fairly simple, and it is related to phase differences. When you have two sound waves that are close in frequency but not quite exact, then the waves would come into phase and go out of phase naturally. This creates a variation in amplitude that we call beats. The frequency of the beat is the absolute value of the difference between the two frequencies.

"Finally, we will discuss standing waves. A standing wave is a wave on a static area, like our hypothetical fixed rope. We can imagine that a wave is going along the rope, sending it up and down. These waves have a certain frequency and a certain wavelength that we can quantify. First, we'll have to define harmonics. Harmonics are the various frequencies a string can hold. For example, the first harmonic, or the fundamental frequency, is the lowest frequency wave that can go on the rope. What is the lowest frequency for a rope that you can think of?"

The class thought about it. At last, A.J. replied, "Well, the fundamental frequency would be when the entire rope just arches up and down."

"Correct. This is the fundamental frequency. The entire length of the spring would be an arc – or an antinode – that goes from one fixed point – or node – to the other fixed point. What is the wavelength of this rope?"

"The wavelength would be twice the length of the rope, because this fundamental frequency is only half of a wavelength," said Todd.

"Very good. And if _v_ _f_ λ, then what would the frequency be?"

"The frequency would be _v_ / 2L, since _f_ is merely 1 / _v_," said Joe.

"Now, if we increase the frequency of the rope, then soon there would be a point when you get an entire wavelength on the length of the string. This is known as the second harmonic."

"The wavelength of the second harmonic would be equal to the length of the rope, and the frequency is _v_ / L," said Brock.

"That is correct. And if we increase the frequency some more, there would be another antinode that appears. This is the third harmonic. There is a basic equation to figuring out the wavelength and frequency of a harmonic. λ 2L / n, where n is a positive integer, and _f_ n_v_ / 2L, where n is the same positive integer. So what would be the λ and frequency of a rope in the 50th harmonic?"

"The frequency would be equivalent to twice the length divided by the harmonic, which is fifty, so the entire expression is (1/25) L. The frequency is equivalent to the harmonic multiplied by the wave speed over the twice the length, so the frequency is 50 times the wave speed over twice the length," answered Samurai.

"Very good, Samurai," said Professor Oak. "Pipes are another situation where you can get standing waves with sound waves. Pipes can have two open ends or one open end and one closed end. The pipe with two open ends is known as an open pipe, and it has the same properties as the fixed rope, except there would be anti-nods on both ends and nodes in the middle. Otherwise the equations you use are the same.

"Closed pipes – or pipes with one closed end – are a monster of a completely different nature. In closed pipes, you have a node on one end and an anti-node on the other end. To figure out which harmonic you are, you would count the nodes, but what would happen in the first and second harmonics?"

"Well, in the first harmonic, there is only one node and one antinode," said Richie, "but you only have a quarter of a wavelength. In the second harmonic, you have two nodes – half a wavelength – but you also have an anti-node beyond that – a quarter of a wavelength. The wavelength becomes (3/4) L."

"Good observation, Richie. This property of closed pipes makes them a little bit more difficult to quantify, but the formula is still very similar. The wavelength is 4L / n, where n is every odd integer. The frequency is n_v_ / 4L, where the n is the corresponding positive integer. This makes it harder to find the wavelength and frequency of the 50th harmonic.

"Well, that's all we have for today. Just remember that this material won't be on the test, so if you don't understand it all, you still have time to get it into your head. We'll see you for the test next class."

The bell rang. Ash was excited that he knew something Gary didn't know. He was going to go brag about it, but his friends stopped him.

"Why do you even want to go talk to Gary when you know he's just going to find a way to make you feel terrible again," said Richie.

"And besides, you have the test to worry about," said Brock.

Ash could only watch as Gary left the room with his cheerleaders. Now, when he finally had the satisfaction of getting one up on Gary, he doesn't even get the opportunity to flaunt it. He sighed and looked at the ground. "This is going to be a long year," he said.

--

AUTHOR'S NOTES: The opening/ending interaction scenes were pretty lame, but I like all of the Pokémon examples I used, even if they were all general examples with a Pokémon makeover. I made up the intensity of the Whismur / Loudred / Exploud sounds, but the Pokédex says that an Exploud's sounds can cause earthquakes…that has to be somewhere like 170 dB, right? You'd think there'd be more Pokémon related casualties in the Pokémon world if they all have some super-strong qualities.


	12. Chapter 12: Light and Optics

Brrring

DISCLAIMER: I don't own Pokémon or the science materials that are in this story. Nintendo owns the Pokémon characters, and the science stuff had better darned well be non-copyrighted. It would suck to have to pay royalties for using the thin lens equation.

--

It was the class after the test. Even though the students had talked about the test almost endlessly after they took it, they continued to do so even before class.

"Man, the multiple choice section was so hard," A.J. complained. "Not only do we have to know how to use the concepts, but we also had to rely on those stupid passages."

"But at least the passages gives us an idea of how these concepts apply to Pokémon training," said Richie. "It makes it seem like it's not just something we learn in class."

"I thought that the free response sections were the hardest, since they were all such complex calculations," said Joe.

"What are you talking about? That was the easiest section," Tracey cried indignantly.

"That's because you're good at those tedious calculations," Todd moaned. "The thing about the free response section is that you can't do any of the rounding like making 9.8 m/s2 equal to 10."

Ash didn't want to hear about the test anymore. He may have pulled an all-nighter cramming the material with Richie, but he just couldn't answer too many questions on the test, especially with the free-response questions. He sneaked a look at Gary, who only wore the confident sneer that was usually on his face. "I hope Gary failed the test too," he muttered.

"I don't think it's good to compete with other people," Brock told him. "It means you have a goal outside of what your true goal should be, and that is to learn."

"And besides," Misty added, "you know what they say, 'Each one should test his own actions. Then he can take pride in himself, without comparing himself to others.'"

"I've never heard that before," said Ash.

The bell rang, and Professor Oak walked into the room. He was carrying his usual set of papers, which indicated that the students couldn't find out what their grades were quite yet.

"Now, you may be wondered if the tests are ready. Well, unfortunately, I'm not completely done grading them all. And there is one student that hasn't taken the test yet," he said.

"What do you mean? You saw me sitting here and take that stupid test of yours!" cried Jessie.

"Writing 'I don't care' or making disparaging comment like 'There's a Snorlax in the way' isn't exactly the same thing as taking the test. I really wish you'd be able to stop by my office and re-take the test to show your true understanding of the concepts."

"Like that'll ever happen," Jessie replied, rolling her eyes.

"Oh well. I'll hope to get your tests back before the next class. For now, let us march on in our journey of the acquisition of knowledge. Today, we'll be talking about the other major wave – light. Light is a transverse wave with perpendicular magnetic and electric fields, so it can also be called an electromagnetic wave. We'll talk about electric and magnetic fields in detail later, but there are some important properties of electromagnetic radiation to keep in mind. First of all, light rays can have different wavelengths and frequencies, but they all have the same velocity. We've discussed this before. Does anybody still remember? Or has all of you forgotten it after taking the test?"

"The speed of light (c) is equivalent to 3 x 108 m/s in a vacuum," Misty replied.

"That is correct, Misty. We talked about this when we discussed nuclear binding energy, but we'll be using it a lot more in our next class, about atomic phenomena. Anyways, with a wave speed of 3 x 108 m/s, light is much faster than sound, which has a wave speed of only 343 m/s. Light is also transmissible in a vacuum, which is how light from the Sun can reach us. There is an electromagnetic spectrum that shows the various sorts of electromagnetic radiation. If you read the book, you should have noticed that the visible radiation we normally refer to as light has wavelengths of only 400nm to 700nm, or 4 x 10-7 m and 7 x 10-7 m.

"As I said, we'll be making calculations with light in our next class. For now, we'll be looking at three important concepts in the field of optics – reflection, refraction, and diffraction. When light enters another medium, some of it reflects off of it while others go into the medium at a different angle – it refracts. Reflection is simple. It bounces off in the same angle as it was going when it struck. In scientific term, θ1 θ2, where θ is the angle against the normal. It is important to keep in mind that the normal in this case is just like the normal force – it is perpendicular against the medium. This means that if you have a medium with a flat horizontal medium, the angle you want is the angle against an imaginary vertical line.

"Refraction is a bit different. Misty, when you're standing in the pool in the Cerulean City gym, haven't you noticed that your lower body in the pool looks like it's separated from your upper body outside of the pool?"

Misty was taken aback at Professor Oak's sudden reference to her, but she quickly recovered and said, "Yes, this is because the light going in is refracted. As you said, light refracted after going into another medium would change its normal angle. The change in angle is equivalent to n1sin(theta1) n2sin(theta2), where n is the index of refraction, or the speed of light in a vacuum (c) divided by the speed of light in the new medium (v). The index of refraction for air is equal to 1, while the index of refraction of water is equal to 1.33."

This time it was Professor Oak that was speechless. "Yes, that is correct, Misty, and great job at explaining Snell's Law so concisely and completely. When you do problems involving refraction, it is important to draw good diagrams and keep the trigonometric functions in mind. Also, it is important to know that theta would decrease if you go from a medium with a lower n to a medium with a higher n. Similarly, θ increases the other way around. In the latter situation, the theta value would increase until it reaches 90°. When this happens, the light doesn't all of a sudden shoot out perpendicular to the boundary, but it actually reflects back. Sin(theta2) in this case would be equal to 1, so you have n1sin(θ1) n2­­. To find the critical angle where this happens between two media, you just take sin-1 of n2 / n1. Remember, n2 must be smaller than n1 because you can't have a positive sine.

"Before we move onto diffraction, it is important to introduce an important application of reflection and refraction: mirrors and lens. In mirrors, light reflects off of a medium to form an image. In lenses, light refracts through the lens to create an image. In both situations you have an object at a certain distance away and you have an image at a certain distance away. The magnification of the object is the negative of the image distance / the object distance. To understand this, it becomes important to follow the sign conventions. A real object or image would have a positive distance, while a virtual object image would have a negative distance. We rarely encounter virtual objects, so it's safe in this class to deal only with real objects. This way, you can either get a real image or a virtual image. A real image would always have a negative magnification because of our definition, and it will be inverted. A virtual image would always have a positive magnification because of our definition, and it will be upright. These statements may not hold if we have virtual objects.

"Whether you get a real image or a virtual image depends on the focal point. Each mirror and lens has a focal point. The focal length is the distance from the mirror or lens to the focal point. For mirrors, the focal length is equivalent to half of the radius of curvature, or the radius of the mirror if the mirror was a complete sphere. As you might expect, a plane mirror like most of the mirrors we have would have an infinite focal point, because the mirror has no curvature. Plane mirrors always result in virtual, upright images.

"The focal length of the mirror depends on the way it curves. For spherical mirrors, let us assume that the mirror is embedded into the wall and that it's impossible to physically put anything beyond it. Anything that is on the real side of the mirror is real – or positive – while anything that is appears to be on the other side of the mirror is virtual – or negative. Concave mirrors – or converging mirrors – curve towards us. These mirrors have a positive focal length because the focal point would be on the real side of the mirror. On the other hand, convex mirrors – or diverging mirrors – curve away from us. In this case, the focal length would be negative because the focal point would be on the other side of the mirror.

"Once you know the focal length of a mirror, there are two ways that you would be able to tell what the image would be like. The first is to draw it out with ray tracing, and the second is to calculate it out with the thin lens equation. Ray tracing isn't too difficult, but you would have to be very exact in your drawing. When you have a converging mirror, the light rays will strike the mirror and go in towards the focal point. You need to draw two rays from the top of the object: one going parallel to the horizontal axis and curving in towards the focal point once it strikes the mirror, and the other going towards the focal point – or coming from the focal point and touching the top of the object if the object is between the focal point and the mirror – and bounce back horizontally as it hits the mirror. Once you draw the rays, then the point where the two rays intersect is where the image will be."

Tracey carefully sketched the examples. "The image will be small, real, and inverted if the object is beyond the focal point, and the image will be large, virtual, and upright if the object is between the focal point and the mirror," he reported.

"That's very good, Tracey. Be sure to show everybody your sketch after the end of class. For a diverging mirror, the rays will reflect away from the mirror, and you must extrapolate to find the focal point on the virtual side. To do ray tracing for diverging mirrors, you just draw the same two rays. For the ray parallel to the horizontal axis, it bounces away after it strikes, but you should be able to trace the reflected ray back to the focal point. For the other ray, continue to draw it towards the focal point, but have it reflect horizontally after striking the ray. Then you extrapolate the horizontal ray on the other side of the mirror. The point where the rays intersect is the image."

Tracey quickly did this sketch and said, "No matter where I put the object, the image always comes out to be small, virtual, and upright."

"Yes. You should be able to see why after I introduce the equation for lenses and mirrors. The inverse of image distance is equal to the inverse of the focal length minus the inverse of the object distance: 1 / i (1 / f) – (1 / o). What can all of you make of this equation?"

Duplica answered, "Well, the diverging lens is easy. You have a negative focal length, and a positive object distance. That means that the image distance will always be negative and the magnification would always be positive. It's a bit trickier for the converging lens. If the object is beyond the focal length, then the inverse of the object distance is smaller than 1/f, so you have a positive image length whose absolute value is larger than the object length and the image will be smaller. If the object is between the focal point and the mirror, the inverse is larger than 1/f and you have a negative image distance whose absolute value is smaller than the object length, and the image will be larger. It is interesting to note that if you have an object at the focal point, you won't have an image at all."

"Yes, you can try that yourself by using a spoon as a mirror. If you look at yourself through the concave end and move the spoon closer – thereby reducing the object distance – there comes a point when the image becomes infinitely large. This is the spoon's focal point. Thin lenses – lens whose thickness are negligible – are very similar to mirrors. They refract light, but the equation for image distance is still the same, and the relationship between objects and images for convergence and divergence are also the same. Converging lenses have a positive focal point because they direct light into the focal point, while diverging lenses have a negative focal point because they scatter light away, and you can trace the lenses back to the focal point."

"So the side on the opposite side of the object is the positive side, and the side on the same side as the object is the negative side?" asked A.J.

"Yes. If you think about it, in both mirrors and lenses, the positive side is the side where light actually goes rather than the side where light only appears to go due to extrapolation."

"So this means that there are completely new sets of laws for ray tracing?" asked Tracey.

"Yes, but they should still be easy for you, Tracey. When you do ray tracing for lenses, you need to draw the focal points on both sides. For converging lenses, which actually have sides that arch out, you draw one ray that initially starts out horizontal and is refracted towards the focal point on the opposite side, and you draw another ray that starts out going through the focal point on the same side as the object – or starting from the focal point if the object is beyond the lens and the focal point – and becomes horizontal when the object passes the lens. The intersection is where the image would be. For diverging lenses, which have sides that cave in, you draw a ray that starts out horizontal and refracts out in a way that you can trace the ray to the focal length on the same side on the object. Then you draw a second ray that goes from the object and passes straight through the middle. The point where these intersect will be the image."

Tracey did the sketching again. "So an object that is beyond the focal point would have a small, inverted image on the opposite side, the positive side of the lens. An object between the focal point and the lens would have a large, upright image on the same side, the negative side of the lens. And for a diverging lens, you always have a small, upright image on the same side, the negative side of the lens," he announced.

"And the math is the same as for mirrors," said Duplica.

"And you should expect that," said Professor Oak, "since the situation is the same except where the light goes. This becomes more complicated when you have virtual objects. This happens when you have a system with multiple lenses not in contact. In this situation, the image of one lens becomes the object of another lens, and if the image is a virtual image then you have a virtual object for the other lens. An example is with microscopes or telescopes. On the other hand, if the lenses are in contact, then you can take it as one lens with a focal point that is equal to the sum of 1/f of all of the lenses.

"However, in real life it's almost impossible to get a lens where thickness is truly negligible. In these situations you need to think that the lens is two spherical mirrors put back to back. The new index of refraction would be 1 / f (n-1)(1 / r1 – 1 / r2, where r1 is the radius of curvature – twice the focal length – of the side closer to you, and r2 is the radius of curvature of the side away from you. Also, if you have to use this equation, remember that r1 and r2 would be negative in sign, because if you take a close look at a lens, the spherical edge would be opposites.

"A final application of optics is diffraction. Diffraction is when a straight ray of light becomes separated after going through a small opening. You can observe this by putting a lens in front of the opening and refracts to a white wall. You should be able to see a band of light that alternates between light and dark. You can calculate the wavelength of light by measuring the angle difference in degrees in between dark bands with the equation a sin (theta) nλ, where a is the size of the slit, theta is the angle between the central light band and the dark band, and n the number of the dark band with 1 being the dark band closest to the central strip.

"This situation becomes more interesting if you have two slits. Based on what we learned last class, waves slightly out of phase with each other will create regions of higher amplitude and regions of lower amplitude. This will happen in this case as well, and you can end up with a series of bands that go from very bright to completely dark. You can quantify the wavelength as well by using both the maxima and the minima. For maxima – where there is intense light – you use the equation d sin (theta) mλ, where λ is the wavelength, d is the distance between slits, θ is the angle between the normal of the middle of the slit and a certain maxima, and m is the maxima that you are measuring. For minima – regions with no light – the equation becomes d sin (theta) (m + ½) λ. It's (m + ½) because the minima are usually centered between maxima.

"Finally, most of the visible lights we see are unpolarized, where the electric field vectors in the electromagnetic radiation are randomly oriented. It is possible to get them to go in one direction with a polarizer." Professor Oak took a polarizer and held it up to the light. The light instantly grew dimmer. "If you take another polarizer and put it perpendicular to the first dialyzer, look what happens." He did this, and the light behind the polarizer went completely dark. "Can anybody explain this?"

"Well, when light went through one polarizer, it only allows light of a particular orientation through," said Richie. "A second polarizer would only allow light of another orientation. When the polarizers are perpendicular, then no light would go through."

"That is correct. Well, this is about all we have time for today. Tomorrow we will deal with the energy of light and how it pertains to electrons. I should also have your tests ready in the next class."

The bell rang, and the students slowly streamed out of the class. Ash watched as Gary walked out with his cheerleaders. "Somebody I'll get the best of you," Ash muttered to himself. "Someday."

"Uhh, isn't it a good thing Gary's been ignoring you?" asked Brock. "And why don't you just ignore him in return?"

"Because he's Gary! He's my rival!" cried Ash. "But just you wait. Somebody I'll get you!"

--

AUTHOR'S NOTES: Yes, there was no chapter that says "Test 1." I decided not to wrack my brain to think of a test with all Pokémon-related passages or problems. Just imagine that there is, and the questions are much better than what I can come up with. Also, Misty's quote comes from Galatians 6:3 (NIV). Yes, Misty is quoting the Bible. Hey, it fit in with the situation.


	13. Chapter 13: Atomic Phenomena

DISCLAIMER: Nintendo owns Pokémon. I don't make any profits from this story. If I ever tried to get this published into another MCAT study book, it would never get off the ground. And I'd get sued if it did.

The students were nervous as they entered the classroom, and they became even more so after seeing Professor Oak standing in front of the room with a large stack of papers on the desk. These were their tests, and he was going to return them today. Some of the students wanted to get the suspense out of the way early, while others were hoping to delay the inevitable.

Ash was in the latter group. He wasn't excited about finding out his grade, especially considering the taunting that he may get from Gary. "No," he said to himself. "I must think positively! I definitely did better than him!" But then he thought again how not only Gary but maybe even Misty and Brock would laugh at the impossibility of that. He groaned and put his head down on the table.

"Hey, Ash, class is about to start!" cried Misty.

And at that moment, the bell rang, and Team Rocket came rushing in at the last second. His mother had been getting better at keeping them away, although her most effective strategy was not using Pokémon but recruiting them to collect an obscene amount of potting soil and fertilizer.

"Good morning, class. As you can see, I have your tests here with me, but we will wait until after I get through with the lesson before passing them out. Today we will take a closer look at light energy and again at the Bohr model. Who can tell me what the Bohr model is?"

Richie raised his hand. "The Bohr model was Niels Bohr's theory of what an electron orbiting a nucleus looks like. According to him, the electron can only occupy orbitals with a certain energy level."

"That is correct, and today we will learn exactly what those energy levels are. First, we will venture back to thermodynamics and talk about blackbody radiation. Any matter above absolute zero would vibrate, and these vibrations produce electromagnetic radiation."

"Even us?" asked Ash.

"That's right, but as we'll soon find out, we're not hot enough to emit visible light."

"Maybe I'm not hot enough to emit visible light, but I'm sure I'm hotter than everybody else here," cried Gary. His cheerleaders cheered behind him, crying "_Gary's so hot! Gary's so bright! Gary could emit visible light!_"

"It's a good rhyme, but as we'll see, that's not going to be physically possible. Physicists determined that the most ideal radiator would also be an ideal absorber – they'd absorb all the light they encounter. Soon these radiators were known as blackbody radiators, because they'd appear black because they're such perfect absorbers. One of the physicists working on this problem was a man by the name of Max Planck, and he derived a constant that is still important today, Planck's constant or h. The value of h is 6.63 x 10-34 (J * sec) or 4.14 x 1034 (eV * sec). And from Planck's analysis we come up with a physical law that relates the peak wavelength emitted to the temperature – called Wien's displacement law – that states λpeak * T = 2.90 x 10-3 m*K. The peak wavelength in this equation is the wavelength where more energy is emitted than any other wavelength. So, what wavelength of light do we emit? And how hot must Gary be to emit visible light, which has a wavelength of 4 x 10-7 m?"

"We are about 3.10 x 102 K," says Tracey, "and the peak wavelength is the constant over our temperature, so the wavelength we emit is 9.35 x 10-3 m."

"Yes," said Professor Oak, "that is about in the infrared range, which is why we use infrared detectors to detect people. And how hot must Gary be to emit visible light?"

"If visible light has a wavelength of 4 x 10-7 m, then Gary has to be 7250K, or 6977°C to emit visible light," Tracey reported.

Most of the class laughed, but Ash kept his mouth shut for fear that Gary would make a disparaging comment towards him.

"The total energy being emitted every second in a unit of area is also related to temperature: ET = σ * T4, where σ is a constant equal to 5.67 x 10-8 J/ (s * m2 * K4). Now, you don't have to memorize these constants, but you do have to know how to use the equations. Now, if Gary's temperature increases twofold, how much would his ET change?"

"Sixteen times, since 24 is 16," said Duplica.

"Very good. This is essentially all I have to say about blackbody radiation, but we would still be using Planck's constant. One of the major applications of this is in the Photoelectric effect, which states that light above a certain frequency that strikes a metal in a vacuum would cause electrons to fly off. This is an important concept to know, but it's more important because it allows us to relate light energy to frequency: E = h_f_, where _f_ is the frequency and h is Planck's constant. Now, you should be able to find the energy if you use are given the temperature of an object by using the Wien's law we learned and the c = λ_f_ equation. You should also be able to see that because of Planck's constant, you can only have certain set values of E. This is how light can have particle properties. The particles are known as photons.

"Now, about the photoelectric effect, each metal has a threshold frequency, which is the minimum frequency of light to drive off electrons. The energy related to the threshold frequency h * _f_T is known as the Work function. This is the minimum energy needed to knock off an electron. The maximum kinetic energy of the emitted electrons is equal to the energy of the light beam minus the work function: h_f_ – h_f_T. If the energy is less than the work function…"

"The kinetic energy of the electron would be zero," said Todd.

"Correct, Todd. The important thing to get out of this – other than to answer questions that appear on tests – is that light energy can lead to effects on electrons. This will be important in our next topic: the Bohr Model revisited. As Richie told us, the electrons in the Hydrogen atom can occupy certain energy levels that correspond to increasing energy. He derived the energy to be equal to be -13.6 / n2, where n is the principle quantum number of the energy level, and the energy is in eV, not Joules. So this means that the energy of the electron in the orbital closest to the nucleus is -13.6 eV."

A.J. raised his hand. "Why is the energy negative?" he asked.

"That's a very good question, A.J. We usually assume that a free electron has energy equal to 0 eV. Once it enters into orbit around a nucleus, it actually loses energy, so you end up with negative energy. And as you can also see, the presence of the n2 value means that there can only be certain allowed energy levels. This is important when it comes to the absorption and emission of light by electrons. Electrons can absorb light and jump to a higher energy level. The caveat is that the photon it absorbs has to be equal to the energy difference between the energy levels. So if you want to find the photon that an electron would absorb to lead to a jump in energy levels, you simply find the difference in energy between the two energy levels: -13.6 eV * ([1 / nf2] – [1 / ni2]).

"Each atom has different energy levels, so the absorption of photons can help physicists fingerprint atoms. They would fire a continuous beam of visible light at a sample of an unknown element. The wavelengths corresponding to energy differences would be absorbed by the electrons, and that would lead to a gap in the spectrum called the absorption spectrum. Similarly, once an electron loses energy, it falls down to a lower energy level and emits a photon that has the same energy as the difference in energy levels. This leads to an emission spectrum that can also be used to identify elements.

"Well, that's I have for you on this subject today. I suppose we do have time to get into our next topic – chemical reaction rates – but I think you have something else on your minds." Professor Oak walked to the pile of tests. "You all did fairly well on this test. It wasn't an easy test, but the lowest score was only a 60%, and the average was actually an 83% with a standard deviation of 12, which means that I may not have to curve the test after all like I was planning to. Be sure to keep up the good work." With that, he began passing the tests back.

The reactions of most of the students were that of relief, although a few were obviously disappointed, like Samurai. "I wonder what he got," Ash thought. "Hopefully it's worse than me so I'm not the one with the worst grade in the class." As soon as he finished the thought, Professor Oak returned his test. Ash took a peek and immediately felt his heart sink. Written on the top with red ink and circled for emphasis was the number 60. This means he was the person with the lowest score. This meant that Gary definitely beat him in this test.

He groaned and put his head down on the table. He turned to the left and saw Brock with a 90 on the test with the words "Very good!" next to it. Then he turned to the right and saw that Richie's test had a 97 and "Excellent!" on it. He couldn't see Misty's test, since she quickly put it into her notebook. "She probably didn't do too well either," he thought.

"I've made a printout of the score of everybody in the class, so you can see how you're compared to the other students," said Professor Oak once he finished returning the exams. "I've replaced the names with the student numbers so you won't be able to tell who everybody is," he added after seeing Gary starting to smirk, possibly in anticipation of seeing Ash's score. "I'll be around to answer any questions you may have about the test."

Professor Oak laid out the score report, and the entire class scrambled to get a look. Ash was too slow, so he stood to the side. He looked down on Professor Oak's desk and noticed the class roster:

01: Imite, Duplica

02: Ketchum, Ash

03: Laramie, Lara

04: Miyamoto, Samurai

05: Oak, Gary

06: Rocket, James

07: Rocket, Jessie

08: Rocket, Meowth

09: Sketchit, Tracey

10: Slate, Brock

11: Smith, Joe

12: Snap, Todd

13: Stone, A.J.

14: Taylor, Richie

15: Waterflower, Misty

He made a mental note of a few key numbers, like Gary and Team Rocket. As he waited to get a look, he heard comments from the other students.

"What? Somebody actually got 100%? I oughta punch that nerd in the face," said A.J.

"Somebody got a score of I? I wonder what that means," said Joe.

"Man! Only five people got above a 90%! That means only five people got an A!" said Lara.

"Meowth! How could you have gotten an 81%?" Jessie cried incredulously.

"Yeah? Well it's better than getting an Incomplete," Meowth replied.

"It's not going to be an Incomplete anymore," Jessie replied. "The Boss made me take the stupid test."

"Really? What did he say?" asked James.

"He said if we don't pass the class, then we'll lose our jobs."

"What? Well, I'm glad I'm at the edge of a D-!"

At last the crowd thinned, and Ash was able to sneak a peak at the score distribution chart.

**Student No.: Total Score**

15: 100

14: 97

09: 95

01: 93

10: 90

03: 86

12: 86

08: 81

11: 81

04: 78

13: 78

05: 77

06: 61

02: 60

07: I

So Ash wasn't the person with the lowest grade! Student 07 – that was probably Jessie – had an Incomplete. He wasn't sure what that meant, but he wasn't on the very bottom row. And he wasn't that far away from the second worst. Student 06 – that was probably James – had a score only one point better than his. And Gary? Gary was Student 05, and his score was 77%, the third worst in the class! Ash almost gagged. He put on so much airs, and yet he's still half a standard deviation below the class average. Looking up, he saw that Brock and Richie were two of the five students with an A. Student 01 got a 93%. Ash remembered that was Duplica. Student 09 got a 95%. Ash tried to remember who that was, but he couldn't. It was probably Tracey. Tracey had always been very smart. And the student with the 100% was Student 15. Student 15 was…Misty!

Ash was shocked. Misty, the annoying redhead that was always complaining and getting on his nerves actually got the only perfect score in the class? He looked over at her. She was standing in front of the door talking to Brock and Richie as if nothing special happened. He never could have expected her to be so smart. It definitely puts a lot of stress on him to be such good friends with three of the five best students in the class. How much do they look down on him because he can't seem to do as well as they do? How much does Professor Oak expect from him just because he hangs out with them? Ash groaned and walked towards them. "It's going to be a long year," he thought to himself.

AUTHOR'S NOTES: The atomic phenomena is right before the nuclear phenomena at the end of the physics section in the MCAT book, but while I put the nuclear phenomena section at the beginning of chemistry, this goes after light, because this is all about what light does to electrons. However, the real juicy part of this chapter is…THE TEST SCORES! Yep, you get to see all of my biases. Okay…not really. I'm not exactly a big fan of Tracey, but he did have that scene in the second movie where he did the calculations real fast, so you know he'd get a good score. But Misty getting a 100% is pure bias right there. The bottom three is pretty much lifted from the episode "The Ultimate Test," with James and Ash getting the lowest scores, and Jessie getting an INCOMPLETE! Actually, Jessie got a 0%, but Professor Oak is kinder than the test administrator. And Gary getting the lowest score by somebody that's not Ash or Jessie or James is also due to the fact I don't like Gary too much. I figured I might as well make him full of hot air or something.


	14. Chapter 14: Kinetics and Equilibrium

DISCLAIMER: Pokémon and all of its characters are the property of Nintendo. I just use it to write stupid stories to help review the sciences for the MCAT.

Ash was still somewhat suspicious of his friends because of their high test scores as he walked into the next class. His behavior towards them had been kind of cold in the days since the last class, and being bright and perceptive people, they've noticed.

"You've been acting kind of different over the past few days," said Richie. "Is everything all right?"

"Never felt better! Everything's fine!" Ash replied.

"Are you sure? You've been kind of testy," said Misty.

Ash was incredulous. "Who's being testy? Not me!"

"Whoa, Ash, just chill," said Brock.

"I am chilled! Look at me! Isn't this just the face of somebody who's chill?" Ash was almost shouting when he finished.

"I only see somebody who's delusional, heh heh heh," said Gary.

"Shut up! Nobody asked you, you pompous freak!" Ash cried back.

"Who are you calling a pompous freak?" Gary demanded.

"You, you idiot," Ash screamed. "You act all tough and stuff but in the end all you got was a 77% on the test!"

Gary was taken aback. He was silent for a few seconds before he went right back on the offensive. "Oh yeah? I bet it was you that got that 60%, which means I'm still one and a half standard deviations better than you!"

"Well, can't say he's wrong about that," Brock said to Misty and Richie.

"Oh yeah? That's still way below the class average! In fact, you ended up…23 points worst than MISTY!" cried Ash.

Most of the rest of the class turned to stare at Misty, who was turning a bright red. Then they started murmuring amongst each other. Misty went over to Ash and grabbed him by the ear. "All right, that's enough frivolous comments from you!"

Gary sneered. "I bet you're just making all that up. You're probably just making all that stuff up to make Misty look good just because you have a MASSIVE CRUSH on her!"

"Shut up, Gary!" said Ash. "You're the one with the massive crush on Misty."

"That's crazy! Why would I like Misty? She's loud, she's obnoxious, she has no manners, and she follows you around like a _(censored)_ following her beloved dog around!" Gary continued on for about a minute or two, talking about the unspeakable acts she committed with Ash and about how ugly she is. Gary finished his diatribe with "Only a complete and utter loser like you could have a massive crush on a girl like that!"

Misty's eyes were full of tears as she struggled to maintain her composure after the vicious verbal attack. In the past she wouldn't have thought twice to pick up a chair and hurl it at Gary, but she's matured over the years and has come to consider such retaliation foolish. Violence only begets violence. She wanted to run out of the room, but that would be a sure sign of weakness. She was clenching her fists so hard that Brock and Richie were afraid that she'd break a vein. "It's all right," Richie said to her, "just remember that it's only Gary."

Misty didn't say a word. She just sat down in her seat and covered her face with her hands. "Gotta stay strong, Misty," she thought to herself, but couldn't stop the tears. She heard some of the students come to her support.

"That was just too mean," she heard Duplica say to Gary. "Misty never did anything to you."

Gary was still defiant to the very end. "Why would I be so mean if all I said were the truth? Besides, she opened herself up to it by hanging out with Ash all the time."

"What are you talking about?" Todd replied. "I find it far more believable that Misty got a 100% on the test than anything you said."

"Misty may be a bit rough around the edges at times," said Tracey, "but she's still a good person. She has the strongest compassion towards people and Pokémon out of anybody I know, and her quick thinking saved our lives several times."

Gary cackled. "Yeah, you losers go around and side with that tramp. Just remember that you'll never be as cool as I am. Well, that would have been impossible anyways…isn't that right, ladies?" Gary's cheerleaders cheered. "_Gary always wins the race! He puts Misty in her place!_" they cried.

The bell rang, and Professor Oak came walking into the classroom. By then Misty had gotten control of her emotions with deep breaths, so Professor Oak started his lesson unaware that a vicious verbal battle had taken place and that his grandson was possibly the most hated student in the class.

"Good morning, everybody," he said. "I hope all of you were satisfied with the first test results, and even if you aren't, just remember that there are three more tests plus a final, so four more opportunities to get your score up. Okay, today we will be changing gears once again and diving back into the fascinating world of kinetics and equilibrium. Kinetics is the study of the rates of a reaction. We study kinetics in order to get a sense of how a reaction mechanism works.

"Reactions are not miraculous coincidences when multiple molecules collide with sufficient energy to drive the reaction. Instead, chemists see it more as a series of steps that ultimately gets you from the reactant to the final product. You may have already known this from Hess's Law in the thermochemistry lesson, where multiple one-step reactions sum up to form a final reaction.

"Here's an example that may help. Suppose you take the reaction of the formation of water: O2 + 2H2 = 2H2O. Instead of thinking that three molecules come colliding at the same time, it may be more helpful to think of the reaction as a two step process. The first step would be the addition reaction between oxygen gas and one mole of hydrogen gas: O2 + H2 = H2O2, and the second reaction would be a single displacement reaction between an oxygen atom on the hydrogen peroxide and another mole of hydrogen gas: H2O2 + H2 = 2H2O. This may not be the true reaction mechanism of water, but it does serve as a simple and helpful picture of a reaction mechanism."

"What then is the true reaction mechanism of water?" asked A.J.

Professor Oak paused. "You know, I don't think that there is one single reaction mechanism, but one of the mechanisms I can think of is the decomposition of two moles of H2O2 to form two moles of water and oxygen gas. We'll deal with more concrete reaction mechanisms later in the course. I won't be testing you on the formation mechanism of water. I will expect you to know that the rate is determined by the slowest step in a multi-step mechanism – the rate determining step.

"The rate of a reaction is the decrease in the concentration of reactants or increase in the concentration of products in a unit of time. The rate has a unit of moles per liter per second. The rate for each reaction is proportional to the concentration of reactants. You can write out a rate law by multiplying the concentration of the reactants together with a rate constant: rate = k [A]x [B]y, where A and B are the reactants while x and y are the orders of a reaction.

"The order of a reaction determines how much the rate will change when you change the concentration of the reactants. For example, if you double a reactant and the rate doubles as well, the order for that reactant is one. If you double a reactant and the rate quadruples, you have an order of two for the reactant, since 4 is two squared. If you double a reactant and the reactant only goes up by the square root of two, then the order for the reactant is one half. The overall order of a reaction is the sum of the order for the reactants. What would happen if you have a reaction with a reaction order of zero?"

Samurai answered, "If the reaction order of a particular reaction is equivalent to the number zero, then the reaction is not dependent on the concentration of the reactants, and the rate will remain the same no matter how much reactant you futilely add."

"That is correct. Chemists usually use these reaction orders to find the mechanism of a reaction. Finding the reaction order is actually quite simple, at least theoretically. You do so by performing a reaction with an exact amount of the reactant and measure the rate. Then you would double the concentration of one of the reactants and measure the effect of the rate. This is difficult in real life because percent yields are usually so low, but the logic behind it is very straightforward."

Joe had a question. "I probably should have asked this earlier, but you said that reactions are not miraculous coincidences with collisions of sufficient between multiple molecules. If they're not dependent on collisions, what are reactions dependent on?"

"Oh," Professor Oak said with a gasp, "I am sorry for being unclear. When I said multiple molecules, I meant more than two molecules. However, reactions are still dependent on collisions, but not all collisions result in a reaction. The collision theory states that a collision must be higher than the activation energy of the reaction between the two molecules, because scientists propose that there would be a high-energy transition state – or activated complex – in the exact moment of electron exchange. The rate would therefore be equal to the number of total collisions between molecules and the percent of collisions that are effective.

Professor Oak drew a potential energy diagram on the white board. "You've seen this before when Blaine drew it. This shows the energy of the reactants with the energy of the products. This hump over here," he said, pointing to a raised section after the reactants, "is the transition energy. The collisions need to have sufficient energy to get over that hump. So even if a reaction is spontaneous, it could still take millions of years to undergo because it would have such a high transition state."

"So lowering the activation energy would increase a reaction rate?" asked Lara.

"That is certainly one way to do it. That's how catalysts increase reaction rates. Each catalyst does it in different ways, but they all lower the reaction rate. It is important to keep in mind that catalysts only lower the reaction rate. They NEVER do anything with the spontaneity of the reaction – Gibb's free energy doesn't change. Some other ways to increase reaction rate include increasing the concentration of reactants – but only if the reaction order is greater than zero – or increasing the temperature, which increases the likelihood of collisions with sufficient energy. Finally, the medium where the reaction occurs can also affect the rate. We'll be learning about this later in the course. Does anybody have any questions?"

"So…you say that catalysts never change the ΔH or the ΔG of the reaction. Are you sure about that?" asked Gary.

"Yes, I am quite sure," Professor Oak replied.

Gary pressed on. "So what if scientists publish a report saying that they found a catalyst that lowered the free energy, what would you say about that finding?"

"I'll say it's a lot of hooey."

"Yeah? Well that'll be my goal! I, Gary Oak, will eventually find a catalyst that would decrease the ΔG of a reaction!" His cheerleaders cheered, crying "_Gary Oak is going on a spree! He'll find a catalyst that changes ΔG!_"

"Well, good luck with that, but just remember that a catalyst can never change the ΔG. Another thing about catalysts is that they will never be consumed in the reaction."

"Well, I'll find one of those too!" Gary added.

"If the catalyst is consumed over the course of a reaction, then it becomes a reactant. We really should move on. We've been assuming that the reactions have been going into one direction. This is because the reverse reaction is usually very slow, although it depends on the reaction. Can anybody explain why?"

Richie raised his hand. "This is because the transition state is the same in both the forward and the reverse reactions so it would be at the same energy, but the energy of the reactants in the reverse reaction is at such a low energy that the activation energy of the reverse reaction is going to be equal to the activation energy of the forward reaction plus the ΔH of the reaction."

"Very good," said Professor Oak. "However, it is important to know that the reverse reaction is never completely extinguished. There will always be collisions with enough energy to get the reverse reaction going. There comes a point, though, when the forward and reverse reactions go at the same rate. This state is known as chemical equilibrium, which is dynamic. Can anybody explain why?"

"Well, unlike translational or rotational equilibrium, where nothing is moving," said Lara, "both the forward and reverse reactions are going on in chemical equilibrium. It's just that they're going at the same rate, so the concentration of the reactants and products aren't going to change."

"That is correct. We can look at this phenomenon in terms of the rate law. Let's take a simple, one step reaction like 2A ←→ B + C. What would the rate law for the forward and reverse reactions be in this case?"

"The rate law for the forward reaction would be rate = kf [A]2. The rate law for the reverse reaction would be rate = kr [B][C]."

Professor Oak wrote these rate laws on the board. "Okay, so these are the rate laws. When we're at equilibrium, the rates are going to be the same, so we can say that kf [A]2 = kr [B][C]. We'd be able to shift this relationship around to get the rate constants on one side and the concentrations on the other: kf / kr = [B][C] / [A]2. If you notice, the concentration of the products is the numerator, and the concentration of the reactants is on the denominator. The coefficients before each term in the balanced equation become the exponents in the equilibrium constant. This term is known as the equilibrium constant KC – or Keq. Each equation should have its own equilibrium constant at a given temperature.

"When you have a multi-step reaction, you can take the product of the rate constant for each individual forward reaction and divide it by the product of the rate constant for each individual reverse reaction and get a KC where the concentrations of the products are on the top and the concentrations of the reactants are on the bottom."

Brock wanted to confirm an observation. "So any of the intermediates that are performed in each individual step are cancelled out in the equilibrium constant?" he asked.

"That's right, Brock. There are a few things to keep in mind about the equilibrium constant. Whenever you have a pure solid or a pure liquid, their concentrations don't appear in the Keq. The value of the equilibrium constant can also tell you about the equilibrium state. Can anybody elucidate?"

Tracey raised his hand. "If the equilibrium constant is a lot larger than one, there would be much more products than reactants, and the forward reaction is favored. If the equilibrium constant is a lot smaller than one, there would be a lot more reactants than products, and the reverse reaction is favored. If the equilibrium constant is close to one, the concentrations of reactants and products and the rate constant for both reactions are about equal."

"Very good, Tracey. As I have said in an earlier class, the spontaneity of a reaction can be related to the equilibrium constant. Does anybody remember how?"

Most of the class was silent. Misty knew the answer, but she was hurting too much on the inside to say anything. She just wrote the equation on her notes: ΔG = -RT ln(Keq). Ash watched her do it.

"It's okay if you don't remember it. The equation is…"

"Triangle G equals dash RTLN Keq!" cried Ash.

"Wow, that's impressive, Ash, but the technical term is Delta G equals negative RT ln(Keq), where R is the universal gas constant and T is temperature in Kelvin. Can you tell us what value for R you would use, or how this relates to spontaneity, Ash?"

"Uh…no"

"Hah! I knew you didn't know what you were talking about," laughed Gary.

"You would use 8.3145 J/(K*mol) for R," said Richie, "and if you have a reaction with a large equilibrium constant, the natural log of that is going to be positive, so the free energy will be negative because of the negative in the equation. This makes sense because at equilibrium there will be much more products than reactants. On the other hand, if the reaction has an equilibrium constant between 0 and 1, the natural log will be negative, and the negative in the equation will make the free energy positive. The reaction will be non-spontaneous, and the system will be mostly reactants at equilibrium."

"Wait," said A.J., "you said earlier that there could be spontaneous reactions that take years to complete. Well, in that case there would be a lot more reactants than products, so Keq would be negative, and wouldn't that make the reaction spontaneous?"

"That's a good observation, A.J., but the problem in your situation is that the reaction has not yet reached equilibrium, so you can't use the equilibrium constant in this case. There is another value for a reaction that you would use for a reaction that hasn't gone to equilibrium. It's called the reaction quotient QC, and it's the concentration of the products over the concentration of the reactants at a particular point in time. At equilibrium, the QC would equal KC, but if the QC is greater, there would be more products than there would be at equilibrium, and if the QC is lower, there would be more reactants than there would be at equilibrium.

"Whenever you apply a stress to a system in equilibrium, the system would be driven out of equilibrium, and it would strive to reverse the change and go back to an equilibrium state. This is known as Le Chatelier's Principle after the chemist who found this principle, and we will use it often in many different sorts of systems. It's important to know that you can look at a reaction and determine whether or not Le Chaelier's Principle would apply.

"There are three major stresses. The first is pretty obvious – it's a change in concentration. If you add reactants to a system in equilibrium, the reaction quotient will fall below one, and the rate of the forward reaction will increase to get the system back into equilibrium. When the rate of the forward reaction increases, we say that the equilibrium is shifted to favor the products. On the other hand, when you increase the concentration of the products, the rate of the reverse reaction will increase, and the equilibrium is shifted to favor the reactants. Of course, changes in concentration don't necessarily have to involve addition of chemicals. What else could we do to change the concentration?"

"You could take away one of the chemicals," said A.J. "If you take away products, the equilibrium will shift to favor the products. If you remove reactants, the equilibrium will shift to favor the reactants."

"Very good. It is important to keep this in mind when thinking about Le Chatelier's Principle. Another stress is a change in pressure. Changes in pressure would affect gases in the system. When thinking about changes in pressure and Le Chatelier's Principle, it's important to think about Boyle's Law."

"That law states that pressure and volume are inversely related," said Lara.

"Correct. The second stress is a change in temperature. When you change the temperature, you have to consider whether or not the reaction is exothermic and endothermic. If you decrease the temperature, then the equilibrium would shift to favor the side where heat is the product, so it would shift towards the reactants in an endothermic reaction and towards the products in an exothermic reaction. On the other hand, if you increase the temperature, the equilibrium would shift away from the side with heat, so it would favor the products in an endothermic reaction and the reactants in an exothermic reaction.

"The final stress is a change in pressure. Whenever you have a situation with a change in pressure, you have to consider the moles of gases on each side, so it is important to balance the equation correctly. If you increase the pressure, the volume will decrease, and the system will shift towards the side with less molecules of gas in order to reduce the stress of the increased pressure. On the other hand, if you decrease the pressure, the volume will increase, and the system will actually shift towards the side with more molecules.

"Le Chatelier's Principle is one of the most important concepts we will learn in chemistry, and it may appear in places where you'd never dream of, so be sure to understand it through and through. This is about all the material we have today." The students began streaming out of the class, many of them giving an evil eye to Gary as they left. When Team Rocket went to get out, Professor Oak stopped them and handed Jessie her test. "Thank you for deciding to retake the test after all. It's a good thing you did because you have tremendous potential."

James and Meowth crowded around Jessie to see what she got. "What? You actually got an 86%?" Meowth exclaimed.

"Yeah," Jessie grumbled.

"So what made you decide to take the test after all?" James asked.

"The boss actually called me and told me that I have to take this class seriously if I still wanted to be a part of Team Rocket…stupid ingrate, threatening me after everything my mama did for Team Rocket. And he did say something along the line that the better we do in this class, the more he'll be willing to overlook the fact that we still haven't caught Pikachu."

"What? That means I'd have to step it up!" James cried sadly.

Meanwhile, Ash, Misty, Brock, and Richie discussed what exactly to do about Gary. He was becoming more and more audacious, asking Professor Oak such stupid questions in class and committing the worst offense of all – insulting Misty.

"Whatever we do," said Brock, "we can't let him get away with it any longer."

"We should probably tell Professor Oak about what he did before class and leave it at that," said Richie.

"I think it's best to just forgive and forget," said Misty.

Ash and Brock were aghast. "After all that he's done to you?" Brock exclaimed. "When did you pick up such Christian values?"

"Whatever. I don't think Professor Oak has any control over Gary, so we'd have to take matters into our own hands," said Ash. He looked towards the door that Gary walked through earlier. "Someday, I'll get you, Gary, if it's the last thing I do!"

AUTHOR'S NOTES: Okay, the entire thing with Gary insulting Misty is probably the most notable part of the chapter, even more notable than the actual science-related stuff. And like all of the other interactions, it was something that I wrote on a whim. The only thing I planned was the conversation between Ash and his friends, which I deliberately made similar to "Pikachu's Goodbye," but then Gary got into the mix, and everything just escalated. I decided to include a reference to Pokéshipping, even though I'm not in favor of it, and Ash's lame comeback was the first thing I came up with. That was probably insulting enough for Misty, but then I decided to go ahead with a more scathing retort by Gary. Misty may be my favorite character in all of Nintendo, but the criticisms were easy to come up with, since I've been hearing it from a bunch of Misty bashers. And then Misty's reaction was another dilemma. I'm sure Misty in the anime wouldn't hesitate to pick up a chair and throw it at Gary before he even got anywhere, but I imagine Misty as an ultra-noble character who would rather take the personal attacks in stride as a sign of strength rather than use retaliation that's a sign of weakness – a true Christian in the sense. I don't know what people's opinions on Misty being a Christian would be, but it's good to think of it that way. Anyways, the transition to the class is horrible, but the story is about MCAT sciences, not the relationships between the characters in the story.


End file.
